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Раздел Курсовые работы
Предмет Алгебра

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Mohigul Xolyigitova

Дата регистрации 25 Октябрь 2024

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Bir noma'lumli ko‘p hadli tenglamalar

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Oliy ta'lim, fan va innovasiyalar vazirligi Jizzax davlat pedagogika universiteti Sirtqi bo`lim Matematika va Informatika yo`nalishi “Tabiiy va aniq fanlarda masofaviy ta’lim” kafedrasi KURS ISHI MAVZU: Bir noma'lumli ko‘p hadli tenglamalar Bajardi: Matematika va Informatikayo`nalishi 3-kurs talabasi Bozorboyeva Dinora Kurs ishi rahbari: Xolyigitova Mohigul Jizzax- 2025 1 R E J A (M U N D A R I J A ) Boblar va paragraflarning nomlanishi betlar Kirish 3 Birinchi bob :Ko`pxadlarni nazariy asoslari 1.1 Birhadlar va ko`phadlar . 5 1.2 K o` p h adlar ustida amallar . 7 1.3 Bеzu tеorеmasi va uni algеbraik kasrlarni soddalash- tirishga tatbiqi . 12 Ikkinchi bob: Bir noma'lumli ko‘p hadli tenglamalar nazariy asoslari 2.1 K o ‘ p h a d l a r h a q i d a u m u m i y t u s h u n c h a . . 16 2.2 K o ‘ p h a d l a r n i n g a s o s i y x o s s a l a r i . 18 2.3. K o ‘ p h a d l a r v a u l a r u s t i d a a m a l l a r 20 Uchinchi bob:O`zgaruvchili ko`pxadlarni turlari 3.1 B i r o ‘ z g a r u v c h i l i k o ‘ p h a d l a r . 27 3.2 K o ‘ p o ‘ z g a r u v c h i l i k o ‘ p h a d l a r . 29 3.3 M a y d o n u s t i d a k o ‘ p h a d l a r . 30 Xulosa 34 Foydalanilgan adabiyotlar va manbalar ro yxatiʻ 36 Kirish A q l n i c h a r x l a s h u c h u n y o d l a s h d a n 2 K o ‘ r a k o ‘ p r o q f i k r l a s h k e r a k . R e n e D e k a r t Ta’limni tubdan isloh qilishning ajralmas va muhim qismi hisoblangan zamonaviy pedagogik texnologiyalar, interfaol metodlar, amaliyotga joriy qilinishi mumkin bo’lgan yo’l-yo’riqlar yangicha usullar orqali mutaxassislar tomonidan ma’lum tizim asosida o’rgatilishi lozim. Jahon ta’limi tajribasi shundan dalolat beradiki, jamiyat taraqqiyoti ta’limning takomillashishi va taraqqiy qilishi bilan chambarchas bog’liqdir. Mustaqillik yillarida ta’lim tizimidagi o’zgarishlar ta’lim jarayoni tarkibiy qisimlarini yangicha tartibda namoyon bo’lishini taqozo etadi. An’anviy ta’lim qonuniyatlari va tamoyillariga mos holda yangi tamoyillarga amal qilish, ta’lim metodlarining paydo bo’lishi, ta’lim vositalarining takomillashuvi, ayniqsa ta’limni tashkil etishning noan’anaviy shakllari keng ko’lamda joriy etilishi kuzatilmoqda. Mazkur o’zgarishlarning bir qismi mavjud didaktik ta’limotlarning tarkibidan o’sib chiqqan bo’lsa, yana bir qismi dunyodagi ta’lim tizimi rivojlangan mamlakatlar olimlarining pedagogik tajribalariga suyangan holda paydo bo’ladi. Shu o’rinda ta’limning zamonaviy interfaol metodlarining mamlakatimiz ta’lim sohasiga kirib kelishi alohida e’tibor qaratadigan yo’nalishlardan biri ekanligini qayd etish zarur. Mavzuning dolzarbligi: Ko‘p hadli tenglamalarning o‘quv dasturidagi o‘rni va ahamiyati Ko‘p hadli tenglamalar matematik bilimlarning muhim qismi hisoblanadi. Ular algebraik tenglamalarning bir turiga kirib, maktab matematikasida asosiy mavzulardan biri sifatida o‘rgatiladi. O‘quvchilarga ko‘p hadli tenglamaning asosiy Nazariy asoslar : Ko‘p hadli tenglamalar algebraik bilimlarni chuqurlashtirish, mantiqiy fikrlashni rivojlantirish va real hayotiy masalalarni yechishda qo‘llaniladigan asosiy vositadir. Amaliy ahamiyat : Bu tenglamalar iqtisodiyot, fizika, texnika va boshqa sohalardagi ko‘plab muammolarni matematik tahlil qilish uchun ishlatiladi. 3 Ta’lim jarayonidagi o‘rni : O‘quvchilarga ushbu mavzuni o‘rgatish nafaqat ularning nazariy bilimini oshiradi, balki ularda turli muammolarni hal qilish ko‘nikmalarini rivojlantiradi. Bu mavzu kelgusida oliy ta’lim yoki kasbiy faoliyatda matematik bilimlarni qo‘llash uchun zarur zamin yaratadi. Metodikaga kirish: Ko‘p hadli tenglamalarni o‘rgatish jarayonidagi muhim elementlar Ko‘p hadli tenglamalarni samarali o‘qitish jarayonida quyidagi metodik elementlar alohida ahamiyatga ega: Tushunchalarni shakllantirish tushunchalarini oddiy va tushunarli qilib tushuntirish. Masalan, "ko‘p had" va "tenglama" tushunchalarini bog‘lash orqali mavzuning mohiyatini ochib berish. Ko‘rgazmali vositalar : Grafiklar, diagrammalar va dasturiy ta’minotlardan foydalanib mavzuni vizual tarzda tushuntirish. Bu o‘quvchilarning mavzuga nisbatan qiziqishini oshirishga yordam beradi. Amaliy misollar : Nazariy ma’lumotlarni amaliy mashqlar bilan mustahkamlash. Masalan, oddiy va murakkab tenglamalarni yechish, turli vaziyatlarga mos yechimlar topish. Differensial yondashuv : O‘quvchilarning bilim darajasi va o‘rganish uslubiga mos dars tashkil qilish. Masalan, kuchliroq o‘quvchilarga murakkabroq vazifalarni berish, zaifroq o‘quvchilarga esa asosiy tushunchalarni mustahkamlash imkoniyatini yaratish. . Maqsad va vazifalar: O‘quvchilarning ko‘p hadli tenglamalarni o‘rganishda chuqur bilim olishini ta’minlash Mavzuni o‘qitishda quyidagi maqsad va vazifalar belgilab olinadi: Maqsad : O‘quvchilarda ko‘p hadli tenglamalar haqida bilim va ko‘nikmalarni shakllantirish, ularda analitik fikrlash va mustaqil masalalarni yechish qobiliyatlarini rivojlantirish. Vazifalar : Nazariy bilimlarni yetkazish : O‘quvchilarga ko‘p hadli tenglamalarning asosiy qonuniyatlari, xususiyatlari va ularni yechish usullari haqida ma’lumot berish. 4 Amaliy ko‘nikmalarni rivojlantirish : O‘quvchilarning mustaqil ravishda masalalarni yechish, xatolarni aniqlash va ularni tahlil qilish qobiliyatini rivojlantirish. Qiziqishni oshirish : Dars jarayonida mavzuni hayotiy misollar bilan bog‘lash orqali o‘quvchilarning mavzuga nisbatan qiziqishini orttirish. Innovatsion yondashuvlarni qo‘llash : Zamonaviy o‘qitish uslublari va texnologiyalaridan foydalangan holda o‘qitish samaradorligini oshirish. Kurs ishiningimizning asosiy ilmiy – uslubiy yangiligi : Ko`pxadlar ustida, uslubiy izlanishlar olib borildi. muammoni yechimlarini o’rganishni takomillashtirish bo’yicha alohida metodik ishlanma ishlab chiqildi. Kurs ishining metodlari: Kurs ishining muammosiga oid pedagogik - psixologik va metodik adabiyotlar mazmunini o’rganish, nazariy jihatdan tahlil qilish, sinflarda o’qitishda zamonaviy texnologiyalarni qo’llash holatini o’rganish. 1 . 1§. Birhadlar va ko`phadlar Birhad dеb, bеrilgan ratsional ifodada katnashuvchi harf ustida ikki amal, ko`paytirish va darajaga ko`tarish natijasida hosil bo`lgan ifodaga aytiladi. Masalan: 2а; 3аbc ; 13 abc 12 ; 3 5 ab ; a 2 b ; xy 4 va h.k. Bеrilgan birhadda ko`paytirishni daraja bilan almashtirib, dastlab o`zgarmas sonni, sungra unda qatnashgan harflarni tеgishli tartibda yozilsa, hosil bo`lgan ifodaga birhadning standart ko`rinishi dеyiladi. Harflar oldidagi sonli ko`paytuvchiga birhadning koeffitsiеnti dеyiladi. 5 Masalan:3abc ⋅5ac ⋅ 2 13 bc bir h adning standart shakl i 30 13 a2b2c3 b o` ladi. Ikki yoki undan orti q bir h adlarning yi g` indisiga k o` p h ad dеyiladi. Dеmak, ko`phad bu birhadlarning algеbraik yig`indisidan iborat bo`lar ekan. Faqat koefitsiеntlari bilan farq qiladigan birhadlarga o`hshash birhadlar dеyiladi. Masalan: 5ab va − 3ab yoki 15 x2у3 va 7х2у3 o`hshash barhadlar, chunki koeffitsiеntlari har xil bo`lib, harfiy ifodalar bir xildir. Ko`p masalalarni yеchishda ikki ko`phad qachon o`zaro tеng bo`ladi dеgan savol tug`iladi. Bu savolga quyidagi tеorema javob bеradi. Tеorеma 1: Agar ikki ko`phadda x ning mos dararjalari oldidagi koeffitsiеntlar tеng bo`lsa, bunday ko`phadlar o`zaro tеng bo`ladi. Masalan: P 2(x)= 3 x2− 7 x+ 4 va Q 2(x)= Ax 2+ Bx +C ko`rhadlarda P2(x)= Q 2(x) bo`lishi uchun А=3; В=-7; С=4 bo`lishi kеrak. Bu tеorеmani qo`llanilishiga bitta misol kеltiramiz: P 3(x)= x3+ 3 x+ 4 uchinchi darajali ko`phadni bitta birinchi va bitta ikkinchi darajali ko`phadlar ko`paytmasi sifatida ifodalash kеrak bo`lsin. Dеmak,birinchi va ikkinchi darajali ko`phadni quyidagi ko`rinishda ifodalaymiz: x+ A va Bx 2+ Cx + D masala shartiga ko`ra x3+ 3x+ 4= (x+ A )(Bx 2+ Cx + D ) bo`lib, tеnglikning o`ng tomonidagi qavsni ochib chiqamiz va x ning bir xil darajalari oldidagi koeffitsiеntlarni tеnglashtiramiz. x3+3x+4= Bx 3+Cx 2+Dx +ABx 2+ACx +AD yoki x3+3x+4= Bx 3+(C +AB )x2+(D +AC )x+AD yoki x3+ 0⋅x2+ 3⋅x+ 4= Bx 3+(C + AB )x2+ (D + AC )x+ AD tеnglashtirsak: B= 1, A= 1, C = − 1, D = 4 ekanligini topamiz. Dеmak, x3+3x+4= (x+1)(x2− x+4) bo`ladi. 6 Tеorеma 2: Agar ikki ko`phadni ko`paytmasi aynan nolga tеng bo`lsa, u holda bu ko`phadlardan hеch bo`lmasa bittasi nolga tеng bo`ladi. 1. 2§. Ko`phadlar ustida amallar Ko`phjadlarni qo`shish uchun ularning har bir hadini o`z ishoralari bilan yozib, hosil bo`lgan yig`indida o`hshash hadlarni ihchamlashtirish kеrak.P = 5 x+ 3 y2− 5 , Q = 5 y2− 4 x+ 4 y , P + Q = (5 x+ 3 y2− 5)+(5 y2− 4 x+ 4 y)= 5 x+3 y2− 5+ 5 y2− 4 x+4 y= ¿(5 x− 4 x)+(3 y2+5 y2)+ 4 y− 5= x+ 8 y2+4 y− 5 . Ko`phaddan yoki birhaddan ko`phadni ayirish uchun kamayuvchining yoniga ayriluvchining hamma hadlarini qarama-qarshi ishora bilan yozib, o`hshash hadlarni ihchamlashtirish kеrak. Misol: P−Q=(5x+3y2−5)−(5y2− 4x+4y)=5x+3y2− 5−5y2+ +4x−4y=9x−2y2− 4y−5 Birhadni ko`phadga ko`paytirish uchun birhadni ko`phadning har bir hadiga ko`paytirib, hosil bo`lgan ko`paytmani qo`shish kеrak. Misol: (3a2b)(5a3b−3abc 2+3 5 ab 3)= 3a2b⋅5a3b− 3a2b⋅3abc 2+ +3a2b⋅3 5ab 3=15 a5b2−9a3b2c2+9 5 a3b4 Ko`phadni ko`phadga ko`paytirish uchun birinchi ko`phadning har bir hadini ikkinchi ko`phadning har bir hadiga ko`paytirib, hosil bo`lgan ko`paytmalarni qo`shish kеrak. Misollar:1. (3a+3b)(a2− 3bc +2ac 2)= 3a⋅a2− 3a3bc +3a⋅2ac 2+3b⋅a2 - − 3b⋅3bc +3b⋅2ac 2 = 3a3− 9abc +6a2c2+3a2b− 9b2c+6abc 2 . 2. (2a−3b+c2)⋅(3a2+2bc +4ac 3)=6a3+4abc +8a2c3−9a2b−6b2c= −12 abc 3+3a2c− 4ac 5 3. (a− b)⋅(a2+ab +c2)= a3+a2b+ab 2− a2b−ab 2− b3= a3− b3 7 Bir h adni bir h adga b o` lish uchun q uyidagi ishlar bajariladi : - B o` linuvchining koeffitsiеnti b o` luvchining koeffitsiеntiga b o` linadi ,h osil b o` lgan b o` linma yoniga b o` linuvchidagi h ar bir h arfni b o` linuvchi va b o` luvchidagi shu h arflar k o` rsatkichlarining ayirmasiga tеng k o` rsatkich bilan yoziladi. - Bo`linuvchining bo`luvchida qatnashmagan harflarini o`zgartirmasdan, bo`luvchining bo`linmada qatnashmagan harflari daraja ko`rsatkichini tеskari ishorasi bilan yoziladi. Masalan: 1). (8a4b3c2):(3a2bc )= 8 3 a4−2 ⋅b3−1⋅c2−1= 8 3 a2b2c. 2). (12 a3b4x4c):(3a2b c3)=4 a3−2⋅b4−1⋅x4c1−3= 4a b3x4c−2 Ko`phadni birhadga bo`lish uchun ko`phadning har bir hadini shu birhadga bo`lib, hosil bo`lgan natijani qo`shish kеrak. Misollar: 1) . (5a4b2−7a5b+6ab 4):(4ab )= 5 3a4−1b2−1−7 3a5−1b1−1+6 3a1−1b4−1= 5 3a3b−7 3a4+2b3 2) . (2a+3a3b−7a2b3):(5a3b2)= 2 5a1−3b−2+3 5a3−3b1−2+7 5a2−3b3−2= 2 5a−2b−2+3 5b−1− 7 5a−1b . Qisqa ko`paytirish formulalari va Nyuton binomi Q uyidagi formulalarga q iska k o` paytirish formulalari dеyiladi. 1. (a+ b)2= a2+ 2 ab + b2 -ikki had yig`indisining kvadrati; 2. (a− b)2= a2− 2ab + b2 -ikki had ayirmasining kvadrati; 3. a2− b2= (a− b)(a+ b) -ikki had kvadratlarining ayirmasi; 4. a3+ b3= (a+ b )(a2− ab + b2) -ikki had kublarining yig`indisi; 5. a3− b3= (a− b)(a2+ab + b2) -ikki had kublarining ayirmasi; 8 6. (a+ b)3= a3+3a2b+3ab 2+b3 -ikki had yig`indisining kubi; 7. (a− b)3= a3− 3a2b+ 3ab 2− b3 -ikki h ad ayirmasining kubi. Kеltirilgan 1-7 formulalar ko`phadni ko`phadga ko`paytirish qoidasiga asosan oson isbotlanadi. Misol uchun 1;5;7 -formulalarning isbotini kеltiramiz: 1. (a+ b)2= (a+ b )(a+ b)= a2+ ab + ab + b2= a2+ 2 ab + b2 5. (a− b)(a2+ ab + b2)= a3+ a2b+ ab 2− a2b− ab 2− b3= a3− b3 7. (a − b )3= (a − b )(a − b )2= (a − b )(a 2− 2 a b + b2)= a 3− 2 a2b + a b2− a2b + 2 a b 2− b3= = a 3− 3 a 2b + 3 a b 2− b 3 Endi q is q a k o` paytirish formulalaridan 1 va 6 formulalarni taxlil q ilamiz: 1. (a+ b)2= a2+ 2 ab + b2 bu formulaning o` ng tomoniga e`tibor bеrsak, a2b0,a1b1,a0b2 h adlar h osil b o` lishida a ning darajasi pasayib , b ning darajasi oshib borayotganini k o` ramiz. 2. (a+ b)3= a3+3a2b+3ab 2+b3 (a+ b)4= (a+b)(a+ b)3= (a+ b)(a3+ 3a2b+3ab 2+ b3)= = a4+4a3b+ 6a2b2+4 ab 3+ b4,яъни (a+b)4= a4+ 4a3b+6a2b2+ 4ab 3+b4 Xuddi shu usul bilan (a+b)5; (a+b)6;...;(a+b)n uchun ikki had yi g` indisini darajaga k o` tarish formulasini h osil q ilish mumkin. Bunda koeffitsiеntlar «Paskal uchburchagi» dеb ataluvchi jadvaldan olinadi. n 0 1 9 1 2 3 4 5 6 7 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 Misol (a+ b)6= a6+ 6a5b+15 a4b2+20 a3b3+15 a2b4+6 ab 5+ b6 Agar (a+b)100 ni ochib chi q ish lozim b o `lsa, yoyilmada 101 ta h ad h osil b o `ladi va bu yoyilma koeffitsiеntlarini Paskal jadvali buyicha h isoblash q iyin b o `ladi. Shu sababli n b a ) (  ni k o `p h adga yoyganda h osil b o `ladigan ak⋅bn−k h ad oldidagi koeffitsiеnt C n k -dan, ya`ni n elеmеntdan k tadan q ilib tuzilgan gruppalashlar sonidan iborat ekanligi isbotlangan, bu еrda C n k= n! k!⋅(n− k)! , n!= 1⋅2⋅...⋅n . Misol: C5 2;C9 5;C12 7 h isoblansin: C 5 2= 5! 2!⋅(5− 3)! = 2!⋅3⋅4⋅5 2!⋅2! = 3⋅4⋅5 2 = 30 ; C 9 5= 9! 5!⋅4! = 5!⋅6⋅7⋅8⋅9 5!⋅1⋅2⋅3⋅4 = 125 C 12 7 = 12 ! 7!⋅(12 − 7)! = 7!⋅8⋅9⋅10 ⋅11 ⋅12 7!⋅1⋅2⋅3⋅4⋅5 = 8⋅3⋅3⋅5⋅2⋅11 ⋅12 8⋅3⋅5 = 6⋅11 ⋅12 = 792 Endi umumiy holda mat е matik induktsiya usuli yordamida N ` yuton binomi d е b ataluvchi quyidagi formulani isbotlaymiz : (a+ b)n= C n 0an+C n 1an−1b+C n 2an−2b2+...+ C n kan−kbk+...C n nbn (1) 10 Bu еrda Cn k -lar binom koeffitsiеntlari dеyiladi va quyidagicha hisoblanadi. C n k= n! k!⋅(n− k)! , C n 0= C n n= 1 , n=1 bo`lsa, (a+ b)1= C 1 0a1+C 1 1b1= a+ b Endi (1) formula n= k bo`lganda o`rinli dеb, uning n= k+1 bo`lganda ham o` rinli ekanligini isbotlaymiz, ya`ni (a+b)k= C k 0ak+C k 1ak−1b+...+C k lak−lbl+...+C k k−1ab k−1+C k kbk (2) bo`lganda (a+b)k+1= C k 0ak+1+C k 1akb+...+C k lak−lbl+...+C k k−1ab k+C k kbk+1 (3) tеnglikning o`rinli ekanligini isbotlaymiz: (a+b)k+1=(a+b)(a+b)k=(a+b)(Ck0ak+Ck1ak−1b+Ck2ak−2b2+...+C klak−lbl+...+ +C k kbk)= Ck 0ak+1+C k 1akb+Ck 2ak−1b2+...+C k lak−l+1bl+C k kab k+Ck 0akb+Ck 1ak−1b2+...+ +C k k−1ab k+C k kbk+1 (4) bundan esa (a+b)k+1= C k 0ak+1+(C k 1+C k 0)akb+(C k 2+C k 1)akb2...+(C k l+C k 1−1)ak−l+1bl+...+ +(C k k+C k k−1)ab k+C k kbk+1 (5) ravshanki, Ck 0= 1= Ck+1 0 , C k k= 1= C k+1 k+1,C n m+1+C n m= n! (m+1)!(n− m− 1)! +n! m !(n− m )! = ¿n! m !(n− m− 1)! ⋅(1 m+1 +1 n− m )= n! m !(n− m− 1)! ⋅n− m+m+1 (m+1)(n− m ) = (n+1)! (m+1)!(n− m )! = Cn+1 m+1 Oxirgi tеngliklarni hisobga olsak, (5) dan (3) tеnglikni o`rinli ekanligini topamiz. Endi mat е matik induksiya usuli bilan (5) formulani umumlashtiramiz, ya`ni An− B n= (A− B )(An−1+ An−2B + An−3B2+...+ AB n−2+ B n−1) (7) formulani isbotlaymiz: n= 2 bo 'lsa , A2− B2=(A− B)(A+B) 11 (7) tеnglikni n= k uchun to`g`ri dеb, n= k+1 uchun isbotlaymiz, ya`ni (A− B )(Ak+ Ak−1B + Ak−2B 2+...+ AB k−1+ B k)= Ak+1+ Bk+1 (8) ekanini isbotlaymiz. (A− B)(Ak+Ak+1B+Ak−2B2+...+AB k−1+Bk)=(A− B)(Ak+Ak+1B+Ak−2B2+...+AB k−1+Bk)+(A+B)Bk= ¿(A− B)(Ak−1+Ak−2B+Ak−3B2+...+Bk−1)A+(A− B)Bk=(Ak− Bk)A+(A− B)Bk= Ak+1− AB k+AB k− Bk+1= ¿Ak+1− Bk+1 shuni isbot qilish talab etilgan edi . N`yuton binomi formulasini ba`zi bir xossalarini o`rganamiz: (x+a)n= C n 0xna0+C n 1xn−1a+C n 2xn−2a2+C n m xn−mam+ ...+C n nx0an (1) 1. Cn 0,Cn 1,C n 2,...,Cn n larga binominal koeffitsiеntlar dеyiladi. 2. N`yuton binomi quyidagi xosalarga ega: 3. N`yuton binomida hadlar soni n-darajadan bittaga ziyod, ya`ni n+1 ta. 4. Unda qatnashayotgan birhadlarda x bilan a ning darajalari yig`indisi (n−m)+m=n ga tеng. 5. Uning umumiy hadi Cn mxn−mam ga tеng b o` lib, Tm+1=C n m x n−m a m , (m= 0,1,2,...n) k o` rinishda bеlgilanadi. 6. N`yuton binomininig oxirgi h adlaridan tеng uzo q likda turgan h adlar o` zaro tеng, ya`ni Cn m= C n n−m ва Cn 0= Cn n= 1;_ Cn 1= Cn n−1= n! 1!(n− 1)! = n,..., 7. N`yuton binomining barcha binomial koeffitsi е ntlari yig`indisi 2n ga tеng. Haqiqatdan (1) formulada x= a= 1 bўlsa, 2n= C n 0+C n 1+ ...+ C n m+ ...+ C n n 8. N`yuton binomida juft va toq o`rnida turgan binomial koeffitsi е ntlar yig`indisi o`zaro t е ng va qiymati 2n−1 ga tеng, ya`ni Cn0+Cn2+Cn4+...= Cn1+Cn3+Cn5+...= 2n−1 1.3§ .Bеzu tеorеmasi va uni algеbraik kasrlarni soddalash- tirishga tatbiqi 12 Bitta o`zgaruvchi x ning ko`bhadi dеb, Pn(x)= anxn+an−1xn−1+⋯ +a1x+a0 (1) ko`rinishdagi ko`phadga aytiladi. аn− kўphadning bosh koeffitsеnti , аn≠ 0 bo`lsa, n soni ko`phadning darajasi а0 - ozod had dеyiladi. Bir o`zgaruvchili ko`phadlar ustida qo`shish , ayrish va ko`paytirish amallari 3 dagi amallar kabi bajariladi. Masalan: P 2(x)=3x2+5x -3, Q3(x)=5x3− x2+7 ko`phadlar bеrilgan P2(x)+Q3(x), P2(x)−Q3(x), P2(x)⋅Q3(x) lar topilsin. Yechish: P 2(x)+Q 3(x)=3x2+5x−3+5x3−x2+7=5x3+2x2+5x+4 P 2(x)−Q 3(x)=3x2+5x−3−5x3+x2−7=−5x3+4x2+5x−10 P 2(x)⋅Q 3(x)=(3x2+5x−3)(5x3− x2+7)= 15 x5− 3x4+21 x2+ +25 x4−5x3+35 x−15 x3+3x2− 21 = 15 x5+22 x4− 20 x3+24 x2+35 x−21 Ko`phadni ko`phadga bo`lish esa xuddi butun sonni butun songa bo`lgani kabi bajariladi, bunda albatta bo`linuvchining darajasi bo`luvchining darajasidan kichik bo`lmasligi kеrak. Bo`lish amalini bajarishda bo`linuvchi ko`phad ham , bo`luvchi ko`phad ham darajalarini pasayish tartibida yozib olinadi, bunda dastlabki o`rinda turgan bo`linuvchining hadi bo`luvchining hadiga bo`linib , bo`linma hosil qilinadi. Masalan: 3х 2 +5х-3 3х 2 -15х Х-5 3х+20 20х-3 20х-100 97 13 yoki 3х2+5х−3 х−5 =3х+20 +97 х−5 Ikkita bir h adning nisbatiga ratsional kasr funktsiya dеyiladi, ya`ni P(x)= Pn(x) Qk(x)= anxn+an−1xn−1+⋯ +a0 bkxk+bk−1xk−1+⋯ +b0 (2) Ratsional kasr funktsiya to`g`ri ratsional kasr dеyiladi, agar n > k bo`lsa va noto`g`ri kasr funktsiya dеyiladi agar n< k bo`lsa. Ravshanki, ratsional kasr funktsiya noto`g`ri bo`lsa, suratini maxrajiga bo`lib, uni bir o`zgaruvchili ko`phad bilan to`g`ri kasrni yig`indisi sifatida ifodalash mumkin. Quydagicha savol tug`iladi. Ko`phadni ko`phadga bo`lish butun sonni butun songa bo`lishga o`hshab kеtmaydimi? yoki 17 3 =5+2 3 yozuv ham noto`g`ri kasrni qoldiqli bo`lishga o`hshamaydimi ? Aslida xaqiqatdan ham P n(x)= anxn+an−1xn−1+⋯ +a0 ko`phad х=0 bo`lgan “n-1” xonali natural sondir. Misollar: 1). 39 =3⋅10 +9 , ах+ b ga (х =10 ) mos k е ladi; 2). 738=7 ¿10 2+3⋅10 +8 , ах 2 +bx+c ga (х=10) mos kеladi; 3). 9675=9 ¿10 3+6⋅10 2+7⋅10 +5 , ax 3+bx 2+c ga ( х =10) ga mos k е ladi. K е ltirilgan misollar qo`yilgan savollarning javobidir. P n(x)=0 ko`rinishdagi t е nglama n- darajali algеbraik tеnglama dеb ataladi. P(x0)=0 b o` lsa, х 0 soni k o` p h adning ildizi dеyiladi. Misol uchun Р 2 (х)=х 2 -8х+15=0 tеnglama uchun х 1 =3, х 2 =5 ildiz bo`ladi, chunki Р 3 (3) =3 2 - 8⋅3+15 = 9− 24 +15 = 0 , Р 2 (5)= 5 2 - 8⋅5+15 = 25 − 10 +15 = 0 XVIII asr oxirida Frantsuz matеmatigi E.Bеzu (1730-1783) quyidagi tеorеmani ta`rifladi va uni isbotladi: Tеorеma:Haqiqiy koeffitsеntli Р n (х) ko`phadni х-а ga bo`lishdagi qoldiq Р n (а) ga tеng. 14 Xususiy holda a soni Р n (х) ko`phadning ildizi bo`lsa ,Р n (x) ko`phad х-а ga qoldiqsiz bo`linadi. Misol uchun Р 2 (х)=3х 2 +5х-3 ko`phadni х-5 ga bo`lganda qoldiq Р 2 (5)= 3⋅52+5⋅5− 3= 97 ga tеng bo`ladi . H a q i q atdan h am 3х 2 +5х-3 3х 2 -15х х-5 3х+20 20х-3 20х-100 97 yoki 3х2+5х−3 х−5 =3х+20 +97 х−5 . Bu tеorеmadan х=а soni Р n (х) ko`phadni ildizi bo`lsa , Р n (х) ko`phadni х-а ga qoldiqsiz bo`linishi kеlib chiqadi . Bu tеorеmani tеskarisi ham o`rinli: Endi Bеzu tеorеmasini algеbraik kasrni soddalashtirishga tatbiqiga misollar kеltiramiz. Ayniqsa, Рn(x) Qk(x) ratsional kasrni soddalashtirishda surat va maxrajdagi ko`phadlarni umumiy x=a ildizga ega bo`lishi kasrning surat va maxrajini x-a ga qisqartirish imkonini bеradi, bundan limitlar nazariyasida 0 0 ko`rinishdagi aniqmasliklarni ochishda foydalaniladi. Misollar: 1). х2−5х+6 х3− х2−14 х+24 kasrni soddalashtiring. Yechish: Kasrni surati Р 2 (х)=х 2 -5х+6, maxraji Q 3 (x)=x 3 -x 2 -14x+24 Bunday h olda q uy i dagi tеorеmadan foydalanish mumkin: Tеorеma: Agar n- darajali (n>1) ko`phadning koeffitsеntlari butun son bo`lib uning ildizi α h am butun son b o` lsa ,u holda α son k o` p h adning b o` luvchisi b o` ladi . Dеmak , amaliyotda bu tеorеmadan foydalanganda ko`phadning ozod hadini butun ko`paytuvchilarga ajratish lozim bo`ladi. 15 P2(x)= x2−5x+6; 6= 2⋅3 , Р 2 (2)=4-10+6=0, Р 2 (3)=9-15+6=0 Р 3 (х)=х 3 +х 2 -14х+24; 24= 8⋅3= 22⋅2⋅3 , Р 3 (х)=64-16-56+24 ¿0 Р 3 (2)=8-4-28+24=0, Р 3 (3)=27-9-92+24=0 D е mak, Р 3 ( х ) ning ozod hadining ko`paytuvchilaridan 4 ildiz emas, 2 va 3 ildiz ekan. B е zu t е or е masiga asosan Р 2 (х), х-2 va x-3 ga q oldi q siz b o` linadi. х 2 -5х+6 =(x-2) (x-3) Ravshanki, bu holda Р 3 (х) ko`phad х2− 5 х+6= (х− 2)(х− 3) ga qoldiqsiz bo`linadi: х 3 - х 2 -14х+24 х 3 - 5х 2 +6х х2− 5 х+6 х+4 4х 2 -20х+24 4х 2 -20х+24 0 Dеmak, х2− 5 х+6 х3− х2− 14 х+ 24 = (х− 2)(х− 3) (х− 2)(х− 3)(х+4) = 1 х+ 4 Bеzu tеorеmasidan amalda qo`llash qulay bo`lgan quyidagi xossalar kеlib chiqadi: 1. Р n(x)= xn− an ko`phad x− a ga qoldiqsiz bo`linadi. Haqiqatdan ham Р n(a)= an− an = 0 2. n= 2k bo`lsa, Pn(x)= xn− an ko`phad x− a ga qoldiqsiz bo`linadi. 3. n= 2 k+1 bo`lsa, Pn(x)= xn− an ko`phad x+a ga qoldiqsiz bo`linadi . 16х 2 -5х+6 х 2 -2х х-2 х-3 -3х+6 -3х+6 0 2 . 1 § . K o ‘ p h a d l a r h a q i d a u m u m i y t u s h u n c h a F a r a z q i l a y l i k , b i z g a .P . m a y d o n b e r i l g a n b o ‘ l s i n . H a r q a n d a y s o n q i y m a t l a r n i q a b u l q i l u v c h i o ‘ z g a r u v c h i ( n o m a ’ l u m ) n i x b i l a n b e l g i l a b q u y i d a g i k o ‘ r i n i s h g a e g a b o ‘ l g a n i f o d a l a r n i h o s i l q i l a y l i k : 1 2 0 1 2 1 ... n n n n n a x a x a x a x a         ( 1 ) B u i f o d a d a 0 1 2, , ,..., n a a a a s o n l a r P m a y d o n n i n g e l e m e n t l a r i d a n i b o r a t b o ‘ l s a , i f o d a n i n g o ’ z i P m a y d o n u s t i d a b e r i l g a n k o ‘ p h a d d e y i l a d i . B u n d a n  m a n f i y b o ‘ l m a g a n b u t u n s o n l a r b o ‘ l i b , 1 2 0 1 2 1 , , , ..., ,n n n n n a x a x a x a x a     K o ‘ p h a d n i n g h a d l a r i , 0 n a x  K o ‘ p h a d n i n g b o s h h a d i , na  e s a k o ‘ p h a d n i n g o z o d h a d i d e y i l a d i , 0 1 2, , ,..., n a a a a  K o ‘ p h a d n i n g k o e f f i t s i e n t l a r i , e s a b o s h k o e f f i t s i e n t d e y i l a d i . ( 1 ) k o ‘ p h a d x n i n g k a m a y i b b o r u v c h i d a r a j a l a r i t a r t i b i d a y o z i l g a n . k o ‘ p h a d n i x n i n g o r t i b b o r u v c h i d a r a j a l a r i t a r t i b i d a h a m y o z i s h m u m k i n : 2 0 1 2 ... . n n a a x a x a x     ( 2 ) B u s o ’ n g i k o ‘ p h a d d a 0a  o z o d h a d v a na  b o s h k o e f f i t s i e n t d i r . ( 1 ) k o ‘ p h a d d a 0 0 a  y o k i ( 2 ) k o ‘ p h a d d a 0 0 a  h o l d a , k o ‘ p h a d n  d a r a j a l i k o ‘ p h a d d e b a t a l a d i . D e m a k , n  d a r a j a l i ( 1 ) k o ‘ p h a d b e r i l g a n b o ‘ l s a , a l b a t t a 0 0 a  , l e k i n q o l g a n k o e f f i t s i e n t l a r d a n b a ’ z i l a r y o k i b a r c h a s i n o l b o ‘ l i s h i m u m k i n e k a n . M i s o l u s h u n : 5 2 5 9 3 7x x x    r a t s i o n a l s o n l a r m a y d o n i d a g i 5 - d a r a j a l i k o ‘ p h a d . S h u n i n g d e k , 6 2 3 2 3 4x x x    h a q i q i y s o n l a r m a y d o n i d a g i 6 - d a r a j a l i k o ‘ p h a d , 2 4 1 (1 3 ) 3 5 ix i x x     e s a k o m p l e k s s o n l a r m a y d o n i d a g i 4 - d a r a j a l i k o ‘ p h a d d i r . 17 E s l a t m a ! K o e f f i t s i e n t l a r i s o n l a r d a n i b o r a t h a r b i r k o ‘ p h a d , e n g o x i r g i n a t i j a d a , k o m p l e k s s o n l a r m a y d o n i d a g i k o ‘ p h a d n i i f o d a l a y d i , c h u n k i k o m p l e k s s o n l a r m a y d o n i o ’ z i c h i g a h a m m a s o n l a r n i o l a d i . K o ‘ p h a d l a r n i q i s q a c h a q i l i b , ( )x  v a ( )x  b o s h q a s h u n g a o ’ x s h a s h s h a k l l a r d a b e l g i l a y m i z . 1 - t a ’ r i f . H a m m a 0 1 2, , ,..., n a a a a k o e f f i t s i e n t l a r i n o l g a t e n g b o ‘ l g a n k o ‘ p h a d n i n o l k o ‘ p h a d d e y i l a d i . D e m a k , b u n d a y k o ‘ p h a d x o ‘ z g a r u v c h i n i n g h a r q a n d a y q i y m a t i d a n o l g a t e n g : ( ) 0f x  . A g a r 0 1 2, , ,..., n a a a a k o e f f i t s i e n t l a r d a a q a l l i b i t t a s i n o l g a t e n g b o ‘ l m a s a , k o ‘ p h a d n o l m a s k o ‘ p h a d d e y i l a d i . S h u n i t a ’ k i d l a s h j o i z k i n - d a r a j a l i n o l m a s k o ‘ p h a d x n i n g b a ’ z i q i y m a t l a r i d a g i n a n o l g a t e n g b o ‘ l i b , q o l g a n q i y m a t l a r n i n g h a m m a s i d a n o l d a n f a r q l i b o ‘ l i s h i h a m m u m k i n ( m a s a l a n , 2 7 12 x x   k o ‘ p h a d 3 x  v a 4 x  q i y m a t l a r d a n o l g a t e n g b o ‘ l a d i ) . A g a r 0 0 a  v a 0 n  b o ‘ l s a , ( )f x k o ‘ p h a d 0 0 0 ( )f x a x a   k o ‘ r i n i s h n i o l a d i v a n o l i n c h i d a r a j a l i k o ‘ p h a d d e b a t a l a d i . D e m a k , P m a y d o n d a g i n o l i n c h i d a r a j a l i k o ‘ p h a d - s h u m a y d o n n i n g n o l d a n f a r q l i e l e m e n t l a r i d a n i b o r a t b o ‘ l a d i . 2 . 2 - § K o ‘ p h a d l a r n i n g a s o s i y x o s s a l a r i . P m a y d o n d a g i i k k i k o ‘ p h a d n i n g a l g e b r a i k t e n g l i g i v a t e n g s i z l i k t u s h u n c h a s i g a t o ’ x t a l a y l i k . 2 - t a ’ r i f . P m a y d o n d a g i i k k i b i r x i l d a r a j a l i : 1 2 0 1 2 1 ( ) ... n n n n n f x a x a x a x a x a          v a 18 1 2 0 1 2 1 ( ) ... n n n n n x b x b x b x b x b          b e r i l i b , k o ‘ p h a d l a r n i n g m o s k o e f f i t s i e n t l a r i t e n g , y a ’ n i 0 0 1 1 2 2 , , ,..., n n a b a b a b a b     b o ‘ l g a n d a g i n a b u i k k i k o ‘ p h a d o ’ z a r o b i r b i r i g a t e n g d e y i l a d i . K o ‘ p h a d l a r n i n g t e n g l i g i ( ) ( )f x x   k o ‘ r i n i s h i d a y o z i l a d i . S h u n i a y t i s h j o i z k i , 0 0 nx  b o ‘ l g a n l i g i u c h u n o ’ z a r o t e n g k o ‘ p h a d l a r b i r - b i r i d a n f a q a t k o e f f i t s i e n t l a r i 0 g a t e n g b o ‘ l g a n h a d l a r i b i l a n f a r q q i l a o l a d i . M a s a l a n , r a t s i o n a l s o n l a r m a y d o n i d a g i : 3 2 ( ) 7 2 1f x x x x     v a 5 4 3 2 ( ) 0 0 7 2 1x x x x x x          K o ‘ p h a d l a r o ’ z a r o t e n g d i r . A g a r ( )f x K o ‘ p h a d n i n g a q a l l i b i t t a k o e f f i t s i e n t i ( )x  K o ‘ p h a d n i n g m o s h a d i k o e f f i t s i e n t i g a t e n g b o ‘ l m a s a , b u k o ‘ p h a d l a r t e n g e m a s h i s o b l a n a d i . M a s a l a n , r a t s i o n a l s o n l a r m a y d o n i d a g i : 8 2 ( ) 3 4 1f x x x x     v a 8 2 ( ) 3 4 7x x x x      K o ‘ p h a d l a r t e n g e m a s . K o ‘ p h a d l a r n i n g t e n g s i z l i g i o d a t d a ( ) ( )f x x   s h a k l d a y o z i l a d i . E l e m e n t a r a l g e b r a d a n m a ’ l u m q o i d a l a r b o ’ y i c h a P m a y d o n d a g i k o ‘ p h a d l a r n i q o ’ s h i s h , a y i r i s h v a k o ‘ p a y t i r i s h m u m k i n . B u n i n g n a t i j a s i d a y a n a P m a y d o n d a g i k o ‘ p h a d l a r h o s i l b o ‘ l a d i , c h u n k i a y t i l g a n a m a l l a r n i b a j a r i s h d a , b i z k o ‘ p h a d l a r n i n g k o e f f i t s i e n t l a r i n i , y a n i P m a y d o n e l e m e n t l a r i n i q o ’ s h a m i z , a y i r a m i z v a k o ‘ p a y t i r a m i z v a s h u s a b a b l i , v u j u d g a k e l u v c h i k o ‘ p h a d l a r n i n g k o e f f i t s i e n t l a r i s i f a t i d a y a n a P m a y d o n e l e m e n t l a r i n i h o s i l q i l a m i z . 19 K o ‘ p h a d l a r n i n g y i g ’ i n d i s i n i n g d a r a j a s i – e n g k a t t a d a r a j a l i q o ’ s h i l u v c h i k o ‘ p h a d d a r j a s i d a n k a t t a e m a s , k o ‘ p h a d l a r k o ‘ p a y t m a s i n i n g d a r a j a s i – k o ‘ p a y t i r i l u v c h i h a d l a r d a r a j a l a r i n i n g y i g ’ i n d i s i g a t e n g . E l e m e n t a r a l g e b r a d a n m a ’ l u m q o i d a g a m u v o f i q k o ‘ p h a d n i n o l m a s k o ‘ p h a d g a b o ‘ l i s h h a m m u m k i n . B o ‘ l i s h q o l d i q l i y o k i q o l d i q s i z b o ‘ l i s h i m u m k i n . B u n d a b o ‘ l i n m a v a q o l d i q h a m y a n a P m a y d o n d a g i k o ‘ p h a d l a r n i i f o d a l a y d i . 2 . 3 - § . K o ‘ p h a d l a r v a u l a r u s t i d a a m a l l a r . B i z b u b o b d a a l g e b r a f a n i u c h u n m u h i m a h a m i y a t g a e g a b o ‘ l g a n k o ‘ p h a d l a r t u s h u n c h a s i b i l a n s h u g ’ u l l a n a m i z . F a r a z q i l a y l i k , b i z g a b i r l i k e l e m e n t g a e g a b o ‘ l g a n b i r o r H b u t u n l i k s o h a s i b e r i l g a n b o ‘ l s i n . 1 - t a ’ r i f .   1, ia H i s    b o ‘ l g a n d a 1 2 1 2 ... sk k k s a a a      ( 1 ) i f o d a H b u t u n l i k s o h a s i u s t i d a b e r i l g a n k o ‘ p h a d d e y i l a d i . B u y e r d a 1 2 ... s k k k    m a n f i y m a s b u t u n s o n l a r b o ‘ l i b , 0 1   v a 1 k k   d e b o l i n a d i . ( 1 ) I f o d a d a u c h r a y d i g a n , ( 1, ) ik i ia x i s   v a  l a r h o z i r c h a b i r o r s i m v o l l a r d e b q a r a l a d i . x s i m v o l o d a t d a n o m a ’ l u m i f o d a d e b y u r i t i l a d i . ( 1 ) i f o d a d a g i ia l a r ( 1 ) k o ‘ p h a d n i n g k o e f f i t s i y e n t l a r i , ( 1, )ik ia x i s  l a r e s a k o ‘ p h a d n i n g h a d l a r i d e y i l a d i . A g a r 0 sa  b o ‘ l s a , sa b o s h k o e f f i t s i y e n t , sk sa x e s a b o s h h a d d e y i l a d i . 20 B i r n o m a ’ l u m l i k o ‘ p h a d l a r o d a t d a       , , . . . f x x q x  o r q a l i b e l g i l a n a d i . k o ‘ p h a d l a r n i n g o ’ z a r o t e n g l i g i , u l a r u s t i d a b a j a r i l a d i g a n a m a l l a r n i q a r a s h d a n o l d i n q u y i d a g i l a r n i t a k i d l a b o ’ t a m i z . 1 . A g a r 1 2 1 ... 0 s a a a      b o ‘ l i b , 0 sa  b o ‘ l s a , ( 1 ) i f o d a d a n sk sa x i f o d a ; 2 . 1 2 1 ... 0 1 1 s s a a a a va k         b o ‘ l s a , ( 1 ) d a n x i f o d a ; 3 . 0 ik   v a 1 2 1 ... 0 s a a a      d a ( 1 ) d a n sa a const   h o s i l b o ‘ l g a n i t u f a y ,sk sa x x v a i s t a l g a n o ‘ z g a r m a s s o n l a r h a m k o ‘ p h a d l a r d e b q a r a l a d i . F a r a z q i l a y l i k ,   f x v a  x  l a r H b u t u n l i k s o h a s i u s t i d a b e r i l g a n ( y a ’ n i k o e f f i t s i y e n t l a r i b u t u n l i k s o h a s i e l e m e n t l a r i d a i b o r a t ) k o ‘ p h a d l a r b o ‘ l s i n . 2 - t a ’ r i f . N o m a ’ l u m n i n g b i r h i l d a r a j a l a r i o l d i d a g i k o e f f i t s i e n t l a r i t e n g b o ‘ l g a n k o ‘ p h a d l a r o ’ z a r o t e n g k o ‘ p h a d l a r d e y i l a d i . 0 0 kx  b o ‘ l g a n i g a b i n o a n o ’ z a r o t e n g k o ‘ p h a d l a r b i r - b i r i d a n f a q a t k o e f f i t s i y e n t l a r i n o l g a t e n g h a d l a r i b i l a n f a r q q i l i s h i m u m k i n . M a s a l a n ,   3 5 f x x x x    v a   2 3 4 5 0 0 0 x x x x x x          K o ‘ p h a d l a r o ’ z a r o t e n g ,   2 4 5 3 h x x x x x     v a   2 4 3 q x x x x    k o ‘ p h a d l a r o ’ z a r o t e n g e m a s . B u t a ’ r i f d a n f o y d a l a n i b , b i z h a r q a n d a y   f x k o ‘ p h a d n i d o i m o q u y i d a g i c h a y o z i s h m u m k i n l i g i g a i s h o n c h h o s i l q i l a m i z : 21   2 0 1 2 ... . n n f x a a x a x a x     ( 2 ) D a r a j a n i n g t a ’ r i f i g a a s o s a n , a g a r 0 na  b o ‘ l s a ,   f x k o ‘ p h a d n  d a r a j a l i d e b y u r i t i l a d i v a u n i n g d a r a j a s i     deg n f x o r q a l i y o z i l a d i , 0a e s a o z o d h a d d e y i l a d i . B a ’ z i h o l l a r d a e s a d a r a j a s i n d a n k i c h i k b o ‘ l g a n  x  k o ‘ p h a d n i n o l k o e f f i t s i y e n t l a r i d a n f o y d a l a n i b , ( 2 ) k o ‘ r i n i s h g a k e l t i r i s h i m i z m u m k i n , y a ’ n i         deg deg k x n f x     b o ‘ l s a ,     1 0 1 0 1 ... ... 0 ... 0 x k k k n k k x b b x b x x b b x b x x                 k a b i y o z i s h m u m k i n . 3 - t a ’ r i f . B a r c h a k o e f f i t s i y e n t l a r i n o l g a t e n g b o ‘ l g a n k o ‘ p h a d n o l k o ‘ p h a d d e y i l a d i . M a z k u r m a z k u r t a ’ r i f g a a s o s a n k a m i d a b i t t a k o e f f i t s i y e n t i n o l d a n f a r q l i k o ‘ p h a d n o l m a s k o ‘ p h a d d e b a t a l a d i . F a r a z q i l a y l i k , n  d a r a j a l i   f x k o ‘ p h a d b i l a n b i r g a l i k d a   deg x s   b o ‘ l g a n   2 0 1 2 ... s s x b b x b x b x       ( 3 ) K o ‘ p h a d h a m b e r i l g a n b o ‘ l s i n . B u n d a y h o l d a i k k i t a   f x v a  x  k o ‘ p h a d n i n g y i g ’ i n d i s i d e b , 22     1 t v v v f x x c x     K o ‘ p h a d n i t u s h u n a m i z . B u y e r d a   max , , v v v t n s c a b   b o ‘ l i b , t s b o ‘ l g a n d a 1 ... 0 s tb b     d e b , t n d a e s a 1 ... 0 n ta a     d e b o l i n a d i . Y a n a s h u n i t a k i d l a y m i z k i , ,v v v va b H a b H     d a y i g ’ i n d i k o ‘ p h a d n i n g d a r a j a s i q o ’ s h i l u v c h i k o ‘ p h a d l a r d a r a j a s i d a n k a t t a e m a s . H a q i q a t a n , a g a r n na b  b o ‘ l s a , y i g ’ i n d i n i n g d a r a j a s i q o ’ s h i l u v c h i k o ‘ p h a d l a r d a r a j a s i d a n h a t t o k i c h i k h a m b o ‘ l i s h i m u m k i n . K o ‘ p h a d l a r t o p ’ l a m i d a a y i r i s h a m a l i o ’ r i n l i . B u t o ’ p l a m d a n o l e l e m e n t s i f a t i d a n o l k o ‘ p h a d q a r a l a d i .   f x K o ‘ p h a d u c h u n q a r a m a - q a r s h i e l e m e n t   2 0 1 2 ... n n f x a a x a x a x      d a n i b o r a t . E n d i x a a x   t e n g l i k b a j a r i l a d i d e b q a r a b , i k k i t a   f x v a  x  k o ‘ p h a d k o ‘ p a y t m a s i d e g a n d a k o e f f i t s i e n t l a r i 0 n s k i k i d a b      t e n g l i k b i l a n a n i q l a n u v c h i k o ‘ p h a d n i t u s h u n a m i z . B u y e r d a 0 0 0 1 0 1 1 0 2 0 2 1 1 2 0 , , , ... d a b d a b a b d a b a b a b       K o ‘ p h a d l a r n i n g k o e f f i t s i e n t l a r i H b u t u n l i k s o h a s i g a t e g i s h l i b o ‘ l g a n i u c h u n 0 na  v a 0 sb  b o ‘ l g a n d a 0 n s n sa b d    b o ‘ l i b ,  0 n n a  v a  0 s s b  d a r a j a l i K o ‘ p h a d l a r k o ‘ p a y t m a s i n i n g d a r a j a s i s h u k o ‘ p h a d l a r d a r a j a l a r i n i n g y i g ’ i n d i s i g a t e n g b o ‘ l a d i . 23 B i z b u n d a n b u y o n n  d a r a j a l i b i r n o m a ’ l u m l i k o ‘ p h a d l a r t o ’ p l a m i n i   H x d e b b e l g i l a y m i z . T e o r e m a . B i r n o m a ’ l u m l i k o ‘ p h a d l a r t o ’ p l a m i   H x b u t u n l i k s o h a s i n i t a s h k i l e t a d i . I s b o t i . I k k i t a k o ‘ p h a d y i g ’ i n d i s i v a k o ‘ p a y t m a s i y a n a k o ‘ p h a d d a n i b o r a t e k a n l i g i n i b i z y u q r i d a k o ‘ r i b o ’ t d i k . E n d i k o ‘ p h a d l a r t o ’ p l a m i u c h u n h a l q a n i n g b o s h q a s h a r t l a r i b a j a r i l i s h i n i k o ‘ r s a t a m i z , c h u n k i b u t u n l i k s o h a s i n i q i s m h a l q a d a n i b o r a t l i g i b i z g a a ’ l u m . 1 . H a q i q a t a n , a g a r a v a b l a r n i y u q o r i d a g i c h a a n i q l a s a k , q u y i d a g i l a r b a j a r i l a d i :   ,v v v v v va b H a b b a     b o ‘ l g a n i u c h u n             0 0 0 0 , t t m n f x x a b x b a x b x a x x f x                                  y a ’ n i k o ‘ p h a d l a r n i q o ’ s h i s h k o m m u t a t i v d i r . 2 .         f x x x f x     ( k o ‘ p a y t i r i s h a m a l i k o m m u t a t i v ) . k o ‘ p h a d l a r n i n g k o e f f i t s i e n t l a r i H b u t u n l i k s o h a s i g a t e g i s h l i b o ‘ l g a n i g a k o ‘ r a 0 0 n s n s k i i k k i i k a b b a          b o ‘ l g a n i t u f a y l i         f x x x f x     b a j a r i l a d i . Y u q o r i d a k o ‘ r i b o ’ t k a n i m i z d e k , 0 na  v a 0 sb  b o ‘ l g a n d a 0 n s n sd a b     . D e m a k ,       0 0 n s n s k i k i k i k i F x f x x a b x d x                 24 K o ‘ p h a d h a m n o l g a t e n g e m a s . D e m a k ,   H x t o ’ p l a m n o l n i n g b o ‘ l u v c h i l a r i g a e g a e m a s . K o ‘ p h a d l a r k o ‘ p a y t m a s i a s s o t s i a t i v d i r , y a ’ n i                 f x x q x f x x q x      ( 4 ) B u t e n g l i k n i i s b o t l a s h u c h u n 0 pc  b o ‘ l g a n d a   2 0 1 2 0 ... c p p m p m m q x c c x c x x c x         d e b o l a m i z .   f x ,  x  v a   q x l a r m o s r a v i s h d a n , s v a p d a r a j a l i b o ‘ l g a n i d a         f x x q x  k o ‘ p h a d d a g i   0,1, ..., ix i n s p    n i n g k o e f f i t s i e n t i 0 0 0 s n p n s p n s k l m k l m j m l k l j k l m i a b c a b c                          Y i g ’ i n d i o r q a l i a n i q l a n s a ,         f x x q x   k o ‘ p h a d d a g i   0,1, ..., ix i n s p    n i n g k o e f f i t s i e n t i e s a 0 n s p p s n s p k i m k l m k l i l m j i k l m a b c a b c                        y i g ’ i n d i o r q a l i a n i q l a n a d i . U l a r n i n g t e n g l i g i g a a s o s a n ( 4 ) t e n g l i k h a m c h i n d i r . 3 .                 f x x q x f x x f x q x      ( 5 ) K o ‘ p h a d l a r n i k o ‘ p a y t i r i s h q o ’ s h i s h a m a l i g a n i s b a t a n d i s t r i b u t i v d i r . B u q o n u n n i n g c h i n l i g i   0 0 0 n s n s n s x k l k l k l k l k l k l a b c a c b c                25 t e n g l i k y o r d a m i d a i s b o t l a n a d i . C h u n k i b u t e n g l i k n i n g o ’ n g t o m o n i        f x x f x q x   y i g ’ i n d i d a g i ix l a r n i n g k o e f f i t s i e n t l a r i d a n , c h a p t o m o n i e s a         f x x q x   d a g i ix n i n g k o e f f i t s i e n t l a r i d a n t u z i l g a n d i r . Y u q o r i d a g i x o s s a l a r d a n q u y i d a g i l a r k e l i b c h i q a d i . 1 .   H x d a b i r n o m a ’ l u m l i b i r n e c h a k o ‘ p h a d l a r y i g ’ i n d i s i t u s h u n c h a s i n i k i r i t i s h m u m k i n . B u n i n g u c h u n i n d u k t i v m e t o d d a n f o y d a l a n a m i z . Y a ’ n i ,   H x h a l q a d a u c h t a       1 2 3 , , f x f x f x k o ‘ p h a d l a r y i g ’ i n d i s i d e g a n d a b i z               1 2 3 1 2 3f x f x f x f x f x f x      n i t u s h u n a m i z . T o ’ r t t a    1, 4 if x i  K o ‘ p h a d y i g ’ i n d i s i h a m a y n a n s h u u s u l d a b e r i l a d i . U m u m a n , n t a   if x k o ‘ p h a d y i g ’ i n d i s i t u s h u n c h a s i h a m q o ’ s h i s h n i n g a s s o t s i a t i v l i g i d a n f o y d a l a n i b k i r i t a o l a m i z , y a ’ n i d a s t a v v a l 1 n t a k o ‘ p h a d y i g ’ i n d i s i   1 1 n i i f x   n i a n i q l a b , u n i n g y o r d a m i d a n t a k o ‘ p h a d n i n g y i g ’ i n d i s i       1 1 1 n n i n i i i f x f x f x            k a b i a n i q l a n a d i . Y u q o r i d a k o ‘ r s a t g a n i m i z d e k , h a r b i r iia x h a d n i k o ‘ p h a d d e b q a r a s h m u m k i n . k o ‘ p h a d a l r n i q o ’ s h i s h a s s o t s i a t i v b o ‘ l g a n l i g i t u f a y l i   2 0 1 2 ... n n f x a a x a x a x      n i n o m a ’ l u m n i n g d a r a j a l a r i n i p a s a y i s h i t a r t i b i d a h a m y o z s a b o ‘ l a v e r a d i . B u n d a y h o l d a   f x k o ‘ p h a d   0, k n ka b k n    a l m a s h t i r i s h y o r d a m i d a   1 2 0 1 2 1 ... n n n n n f x b x b x b x b x b          k o ‘ r i n i s h n i o l a d i . 26   H x d a n o l i n g a n x v a kx l a h a m d a 1 H  u c h u n 1 ; 1 k k x x x x    b o ‘ l g a n i v a k l k l ax bx abx    t e n g l i k k a b i n o a n 2 3 , , ... , nx x x s i m v o l l a r n i x n o m a ’ l u m n i n g d a r a j a l a r i d e b q a r a s h i m i z m u m k i n . H a q i q a t a n ,      2 2 2 1 1 1 1 x x x x x       ,          3 2 2 3 3 1 1 1 1 1 1 x x x x x x x          v a h o k a z o . B u l a r d a n t a s h q a r i , H b u t u n l i k s o h a s i d a n o l i n g a n i s t a l g a n ia H  e l e m e n t n i (   0 0, i ia x x i n   b o ‘ l g a n i t u f a y l i ) n o l i n c h i d a r a j a l i k o ‘ p h a d , ix n i e s a i x t i y o r i y b i r h a d d e b q a r a b , k o ‘ p h a d l a r n i k o ‘ p a y t i r i s h g a b i n o a n iia x b i r h a d n i    0 1 i ii ia x a x x   k a b i y o z i s h m u m k i n . k o ‘ p a y t i r i s h   H x d a k o m m u t a t i v b o ‘ l g a n l i g i t u f a y l i i ii ia x x a  s h a r t h a m b a j a r i l a d i . D e m a k , k o e f f i t s i e n t l a r i H b u t u n l i k s o h a s i g a t e g i s h l i b o ‘ l g a n x n o m a ’ l u m l i k o ‘ p h a d l a r t o ’ p l a m i k o m m u t a t i v h a l q a e k a n . B u n d a n t a s h q a r i , a g a r ( 2 ) v a ( 3 ) d a 0 a v a 0 sb  d e s a k , b u l a r d a n     0 , 0 f x x    b o ‘ l i b , u l a r n i n g k o ‘ p a y t m a s i b o ‘ l m i s h       0 F x f x x    b o ‘ l a d i . C h u n k i       0 0 0 n s n sa b d       . D e m a k ,   H x h a l q a b u t u n l i k s o h a s i d a n i b o r a t e k a n . 4 - t a ’ r i f . A g a r k o ‘ p h a d l a r n i n g k o e f f i t s i e n t l a r i b i r o r P m a y d o n g a t e g i s h l i b o ‘ l s a ,   P x g a P m a y d o n u s t i d a q u r i l g a n k o ‘ p h a d l a r h a l q a s i ( b u t u n l i k s o h a s i ) d e y i l a d i . 3 . 1 - § B i r o ‘ z g a r u v c h i l i k o ‘ p h a d l a r A s o s i y t u s h u n c h a l a r : n - d a r a j a l i k o ‘ p h a d , k o ‘ p h a d n i n g i l d i z i , B e z u t e o r e m a s i , a l g e b r a i k t e n g k o ‘ p h a d l a r , f u n k t s i o n a l t e n g k o ‘ p h a d l a r . 27 A g a r 0 na  b o ‘ l s a , u h o l d a u s h b u 1 1 1 0 0 ... ( , 0,1, ) n n n i n n i i i a x a x a x a a x a K i n N              i f o d a n i K m a y d o n u s t i d a g i n - d a r a j a l i k o ‘ p h a d d e y i l a d i . A g a r K b u t u n l i k s o h a s i n i n g b i r o r c e l e m e n t i u c h u n f ( c ) = 0 t e n g l i k o ’ r i n l i b o ‘ l s a , u h o l d a c e l e m e n t f ( x ) k o ‘ p h a d n i n g y o k i f ( x ) = 0 t e n g l a m a n i n g i l d i z i d e y i l a d i . B e z u t e o r e m a s i . f ( x ) k o ‘ p h a d n i x - s i k k i h a d g a b o ‘ l i s h d a n x o s i l b o ‘ l g a n q o l d i q f ( c ) g a t e n g . x = s e l e m e n t f ( x ) k o ‘ p h a d n i n g i l d i z i b o ‘ l i s h i u c h u n f ( x ) n i n g x - s i k k i h a d g a b o ‘ l i n i s h i z a r u r v a e t a r l i . A g a r 1 2, , ..., k s s s l a r f ( x ) k o ‘ p h a d n i n g t u r l i i l d i z l a r i b o ‘ l s a , u h o l d a f ( x ) k o ‘ p h a d 1 2 ( )( ) ... ( ) k x s x s x s    k o ‘ p a y t m a g a b o ‘ l i n a d i . N o l d a n f a r q l i n - d a r a j a l i k o ‘ p h a d ( n ≥ 1 ) K b u t u n l i k s o h a s i d a n t a d a n o r t i q i l d i z g a e g a e m a s . A g a r     f x K x  v a 0 ≠ φ ( x )  K [ x ] k o ‘ p h a d l a r b e r i l g a n b o ‘ l i b , s h u n d a y g ( x )  K [ x ] k o ‘ p h a d t o p i l s a k i , n a t i j a d a f ( x ) = φ ( x ) g ( x ) t e n g l i k o ’ r i n l i b o ‘ l s a , u h o l d a f ( x ) k o ‘ p h a d φ ( x ) k o ‘ p h a d g a b o ‘ l i n a d i d e y i l a d i v a u n i     f x x   y o k i f ( x ) / φ ( x ) k o ‘ r i n i s h l a r d a b e l g i l a n a d i . O ‘ z g a r u v c h i n i n g b i r x i l d a r a j a l a r i o l d i d a g i k o e f f f i t s i e n t l a r i t e n g b o ‘ l g a n k o ‘ p h a d l a r o ’ z a r o a l g e b r a i k m a ’ n o d a g i t e n g k o ‘ p h a d l a r d e y i l a d i . A g a r o ‘ z g a r u v c h i n i n g b i r o r c h e k s i z s o h a d a n o l i n g a n x a r q a n d a y q i y m a t l a r i g a m o s k e l u v c h i k o ‘ p h a d l a r n i n g q i y m a t l a r i u s t m a - u s t t u s h s a , u h o l d a b u n d a y k o ‘ p h a d l a r n i o ’ z a r o f u n k t s i o n a l m a ’ n o d a g i t e n g k o ‘ p h a d l a r d e y i l a d i . 28 B e r i l g a n 1 1 1 0( ) ...n n n nf x a x a x a x a        K o ‘ p h a d n i 1 1 1 0 ( ) ...m m m mg x b x b x b x b        K o ‘ p h a d g a b o ‘ l i s h n i q u y i d a g i j a d v a l a s o s i d a b a j a r i s h m u m k i n : a n a n - 1 a n - 2 … a m a m - 1 … a 0 b m a n 1 n m m a b b 2 n m m a b b  b m - 1 1 1na    1 1 1 n m m a b b     . . . b 1 b 0 … … n m ma    1 m n m m ma b b     … 0 m m n m a b b     n m n m c a b  2 1 1 n m n m c a b         … … 0 n m m m c a b       1 1 1 m m m d a          … 0 0 0 d a      29 3 . 2 - § K o ‘ p o ‘ z g a r u v c h i l i k o ‘ p h a d l a r A s o s i y t u s h u n c h a l a r : k o ‘ p o ‘ z g a r u v c h i l i k o ‘ p h a d , k o ‘ p h a d n i n g d a r a j a s i , b i r j i n s l i k o ‘ p h a d , l e k s i k o g r a f i k y o z i l g a n k o ‘ p h a d , s i m m e t r i k k o ‘ p h a d , a s o s i y ( e l e m e n t a r ) s i m m e t r i k k o ‘ p h a d l a r , k o ‘ p h a d l a r n i n g r e z u l t a n t i . K a m i d a i k k i t t a o ‘ z g a r u v c h i g a b o g ’ l i q b o ‘ l g a n k o ‘ p h a d k o ‘ p o ‘ z g a r u v c h i l i K o ‘ p h a d d e y i l a d i . n t a n o m a ’ l u m l i k o ‘ p h a d 1 2 ... i i i n x x x   k o ‘ r i n i s h d a g i c h e k l i s o n d a g i h a d l a r n i n g a l g e b r a i k y i g ’ i n d i s i d a n i b o r a t . n t a 1 2, ,..., n x x x o ‘ z g a r u v c h i l i k o ‘ p h a d 1 1 2 1 2 ( , ,..., .. ) . i i i n n i i n f a x x x x x x       k o ‘ r i n i s h d a b o ‘ l a d i . B u n d a ia K  . n t a n o m a ’ l u m l i k o ‘ p h a d n i n g d a r a j a s i d e b , b u k o ‘ p h a d d a g i q o ’ s h i l u v c h i l a r d a r a j a l a r i n i n g k a t t a s i g a a y t i l a d i . B a r c h a k u s h i l u v c h i l a r i n i n g d a r a j a l a r i b i r x i l b o ‘ l g a n k o ‘ p h a d g a b i r j i n s l i K o ‘ p h a d y o k i f o r m a d e y i l a d i . 1 2 ( , ,..., ) n f x x x k o ‘ p h a d b e r i l g a n b o ‘ l i b , u n i n g i k k i t a h a d i d a n q a y s i b i r i d a x 1 n i n g d a r a j a s i k a t t a b o ‘ l s a , u s h a h a d y u q o r i h a d d e b y u r i t i l a d i . A g a r b u h a d l a r d a g i x 1 n i n g d a r a j a s i t e n g b o ‘ l i b , q a y s i b i r i d a x 2 n i n g d a r a j a s i k a t t a b o ‘ l s a o ’ s h a h a d y u q o r i d e b x i s o b l a n a d i v a x . k . 1 2 ( , , ..., ) n f x x x k o ‘ p h a d n i b i r i n c h i o ’ r i n d a e n g y u q o r i h a d n i , i k k i n c h i o ’ r i n d a q o l g a n h a d l a r o r a s i d a e n g y u q o r i b o ‘ l g a n h a d n i v a s h u j a r a y o n o x i r g i h a d u c h u n y o z i l g a n b o ‘ l s a , u h o l d a 1 2 ( , ,..., ) n f x x x k o ‘ p h a d l e k s i k o g r a f i k y o z i l g a n d e y i l a d i . A g a r k o ‘ p n o m a ’ l u m l i k o ‘ p h a d d a g i i x t i y o r i y i k k i t a n o m a ’ l u m n i n g o ’ r i n l a r i n i a l m a s h t i r g a n d a k o ‘ p h a d o ‘ z g a r m a s a , u h o l d a b u n d a y k o ‘ p h a d s i m m e t r i k k o ‘ p h a d d e y i l a d i . 30 1 2, , ..., n x x x n o ‘ z g a r u v c h i l a r d a n t u z i l g a n 1 2 1 1 221 1 1 3 1 .................... ... ... .. ... n n n n n x x x x x x x x x x x x                     s i s t e m a d a g i s i m m e t r i k k o ‘ p h a d l a r a s o s i y ( e l e m e n t a r ) s i m m e t r i k k o ‘ p h a d l a r d e y i l a d i . S i m m e t r i k k o ‘ p h a d l a r h a q i d a g i a s o s i y t e o r e m a . F m a y d o n u s t i d a g i x a r q a n d a y s i m m e t r i k k o ‘ p h a d s h u F m a y d o n u s t i d a g i e l e m e n t a r s i m m e t r i k k o ‘ p h a d l a r o r q a l i y a g o n a u s u l d a i f o d a l a n a d i . 3 . 3 - § M a y d o n u s t i d a k o ‘ p h a d l a r A s o s i y t u s h u n c h a l a r : k e l t i r i l a d i g a n k o ‘ p h a d , k e l t i r i l m a y d i g a n k o ‘ p h a d , A g a r F m a y d o n u s t i d a b e r i l g a n v a d a r a j a s i n o l g a t e n g b o ‘ l m a g a n f ( x ) K o ‘ p h a d n i s h u m a y d o n u s t i d a g i v a d a r a j a l a r i f ( x ) n i n g d a r a j a s i d a n k i c h i k i k k i t a g ( x ) , h ( x ) k o ‘ p h a d l a r k o ‘ p a y t m a s i s h a k l i d a i f o d a l a s h m u m k i n b o ‘ l s a , u h o l d a f ( x ) K o ‘ p h a d n i F m a y d o n u s t i d a k e l t i r i l a d i g a n k o ‘ p h a d , a k s i n c h a , a g a r b u n d a y K o ‘ p a y t m a s h a k l i d a i f o d a l a s h m u m k i n b o ‘ l m a s a , u h o l d a f ( x ) n i F m a y d o n u s t i d a k e l t i r i l m a y d i g a n k o ‘ p h a d d e y i l a d i . A l g e b r a n n i g a s o s i y t e o r e m a s i . D a r a j a s i 1 d a n k i c h i k b o ‘ l m a g a n k o m p l e k s k o e f f i t s i e n t l i x a r q a n d a y k o ‘ p h a d k a m i d a b i t t a k o m p l e k s i l d i z g a e g a . A g a r d ( x ) k o ‘ p h a d f ( x ) v a φ ( x ) k o ‘ p h a d l a r n i n g u m u m i y b o ‘ l u v c h i s i b o ‘ l i b , d ( x ) k o ‘ p h a d f ( x ) v a φ ( x ) l a r n i n g i x t i y o r i y u m u m i y b o ‘ l u v c h i s i g a b o ‘ l i n s a , u h o l d a d ( x ) b o ‘ l u v c h i n i f ( x ) v a φ ( x ) k o ‘ p h a d l a r n i n g e n g k a t t a u m u m i y b o ‘ l u v c h i s i ( E K U B ) d e y i l a d i v a u n i ( f ( x ) ; φ ( x ) ) k o ‘ r i n i s h d a b e l g i l a n a d i . . 1 1 1 1 0 1 1 0 ( ) ... , ( ) ... n n n n n n n n f x a x a x a x a f y a y a y a y a               . 31 b o ‘ l s i n . 1 1 2 1 1( ) ( ) ( ) ( ) ( ... ) ( ) ( ; ) n k k k k n k k k k k f x f y a x y x y a x x y y x y F x y                   b u e r d a 1 2 1 1 ( ; ) ( ... ). n k k k k k F x y a x x y y          A y t a y l i k x = y b o ‘ l s i n . U h o l d a 1 1 1 2 1 ( ; ) 2 ... n k n k n k F x x ka x a a x na x          b o ‘ l i b , F ( x ; x ) n i f ( x ) k o ‘ p h a d n i n g f o r m a l h o s i l a s i d e y i l a d i v a u n i ) '( f x y o k i 'f o r q a l i b e l g i l a n a d i . 1 1 1 0 ( ) ... n n n n f x a x a x a x a        K o ‘ p h a d n i x - s n i n g d a r a j a l a r i b u y i c h a ( ) 2 ''( ) ( ) ( ) ( ) '( ) ( ) ( ) ... ( ) 2! ! n n f c f c f c f c f x c x c x c n f x         k o ‘ r i n i s h d a y o z i l a d i . K o m p l e k s s o n l a r m a y d o n i u s t i d a 1 1 1 0 ( ) ... n n n n f z c z c z c z c        K o ‘ p h a d b e r i l g a n b o ‘ l i b , 1 2, ,..., n    l a r ( )f z K o ‘ p h a d n i n g i l d i z i b o ‘ l s a , u h o l d a u s h b u 1 1 2 2 1 2 1 3 1 3 1 2 3 1 2 4 1 2 1 2 ... ; ... ; ... ; ..................................... ( 1) ... ; n n n n n n n n nc c c c a                                       m u n o s a b a t l a r o ’ r i n l i b o ‘ l a d i . K o m p l e k s s o n l a r m a y d o n i C u s t i d a g i u s h b u 3 2 ( ) 0 0 ax bx cx d a      k o ‘ r i n i s h d a g i t e n g l a m a 3 - d a r a j a l i b i r n o m a ’ l u m l i t e n g l a m a d e y i l a d i . U n i n g x a r i k k i q i s m i n i a g a b o ‘ l i b 32 3 2 0 b c d x x x a a a    t e n g l a m a n i x o s i l q i l a m i z . U n d a a l m a s h t i r i s h b a j a r i b , s o d d a l a s h t i r g a n d a n s o ’ n g 3b x y a   t e n g l a m a n i x o s i l q i l a m i z . B u n d a 3y ry q   a l m a s h t i r i s h d a n s o ’ n g u v a v l a r n i s h u n d a y t a n l a b o l a m i z k i , n a t i j a d a 3 u v + r = 0 b a j a r i l s i n . U h o l d a 3 3 3 3 3 v v 27 u q p u        s i s t e m a g a e g a b o ‘ l a m i z . S i s t e m a d a n k o ‘ r i n a d i k i 3u v a 3v l a r V i e t t e o r e m a s i g a k o ‘ r a q a n d a y d i r 3 2 0 27 p z qz    t e n g l a m a n i n g i l d i z i b o ‘ l a d i . B u k v a d r a t t e n g l a m a n i y e c h i b 3 1z u  d a n 2 3 3 , 2 4 27 q q p u     2 3 3 v 2 4 27 q q p     l a r n i x o s i l q i l a m i z . u v a v n i n g x a r b i r i g a u c h t a q i y m a t , u o ‘ z g a r u v c h i u c h u n e s a t o ’ q q i z t a q i y m a t t o p i l a d i . A g a r 2 , ,u u u E E ( b u n d a E s o n 1 d a n c h i q a r i l g a n 3 - d a r a j a l i i l d i z ) 1z n i n g u c h i n c h i d a r a j a l i i l d i z l a r i n i n g q i y m a t l a r i b o ‘ l s a , u n g a m o s 2z n i n g u c h i n c h i d a r a j a l i i l d i z l a r i q i y m a t l a r i 2 v , v , v Ev E b o ‘ l a d i . N a t i j a d a k e l t i r i l g a n t e n g l a m a 1 v y u   , 2 2 2 3 v, v y u y u     E E E E i l d i z l a r g a e g a b o ‘ l i b , u n d a 1 3 2 2 i   E b o ‘ l g a n i u c h u n 1 v y u   , 2 1 3 ( v) + i ( v), 2 2 y u u    3 1 3 ( v) + i ( v) 2 2 y u u    b o ‘ l a d i . B u e r d a 33 3 x = y - в аn i e ’ t i b o r g a o l i b b e r i l g a n t e n g l a m a n i n g 1 1 3 x = y - в а , 2 2 3 x = y - в а i l d i z l a r i t o p i l a d i . K u b t e n g l a m a n i b u u s u l d a e c h i s h u n i K a r d a n o u s u l i b i l a n e c h i s h d e y i l a d i . A g a r 3 0 x px q    t e n g l a m a d a r , q l a r h a q i q i y s o n l a r b o ‘ l i b , 2 3 4 27q p    b o ‘ l s a , u h o l d a q u y i d a g i m u l o h a z a l a r o ’ r i n l i : 1 ) A g a r 0   b o ‘ l s a , t e n g l a m a b i t t a h a q i q i y v a i k k i t a o ’ z a r o q o ’ s h m a m a v x u m i l d i z l a r g a e g a b o ‘ l a d i ; 2 ) A g a r 0   b o ‘ l s a , t e n g l a m a n i n g b a r c h a i l d i z l a r i h a q i q i y v a k a m i d a b i t t a i l d i z i k a r r a l i b o ‘ l a d i ; A g a r 0   b o ‘ l s a , t e n g l a m a n i n g b a r c h a i l d i z l a r i h a q i q i y v a t u r l i c h a . A g a r a b u t u n s o n k o e f f i t s i e n t l a r i b u t u n b o ‘ l g a n 1 1 1 0 ... n n n na x a x a x a       t e n g l a m a n i n g i l d i z i b o ‘ l s a , u h o l d a ( 1) 1 f a   v a ( 1) 1 f a   s o n l a r h a m b u t u n s o n l a r b o ‘ l a d i . A g a r r / q ( q > 0 ) q i s q a r m a s k a s r k o e f f i t s i e n t l a r i b u t u n b o ‘ l g a n 1 1 1 0 ... n n n na x a x a x a       t e n g l a m a n i n g i l d i z i b o ‘ l s a , u h o l d a r s o n na o z o d h a d n i n g q s o n e s a 0a b o s h k o e f f i t s i e n t n i n g b o ‘ l u v c h i s i b o ‘ l a d i . E y z e n s h t e y n k r i t e r i y a s i . B u t u n k o e f f i t s i e n t l i 1 1 1 0 ( ) ... n n n n f x c x c x c x c        K o ‘ p h a d n i n g b o s h k o e f f i t s i e n t i nc d a n b o s h q a b a r c h a k o e f f i t s i e n t l a r i r t u b s o n g a b o ‘ l i n i b , o z o d h a d 0c e s a 2r 34 g a b o ‘ l i n m a s a , u h o l d a( )f x K o ‘ p h a d r a t s i o n a l s o n l a r m a y d o n i u s t i d a k e l t i r i l m a y d i g a n k o ‘ p h a d b o ‘ l a d i . k o ‘ p K a s r n i n g m a x r a j d a g i i r r a t s i o n a l l i k n i y o ’ q o t i s h m u m k i n , y a ’ n i 1F s o n l a r m a y d o n i u s t i d a k e l t i r i l m a y d i g a n n – d a r a j a l i 1 1 1 ( ) ... n n n n r x x a x a x a        ( n ≥ 2 ) K o ‘ p h a d b e r i l g a n b o ‘ l i b , x = α u n i n g i l d i z i b o ‘ l s a , u h o l d a ( ) ( ( ) 0) ( ) f g g     k a s r r a t s i o n a l i f o d a n i s h u n d a y o ‘ z g a r t i r i s h m u m k i n k i , n a t i j a d a u n i n g m a x r a j i b u t u n r a t s i o n a l i f o d a g a a y l a n a d i . Xulosa O‘qitish jarayonida olingan natijalarni tahlil qilish, o‘quvchilarning bilim darajasini baholash va ko‘p hadli tenglamalar mavzusini o‘rgatishda foydalaniladigan usullarning samaradorligini ko‘rib chiqishga bag‘ishlanadi. Nazorat va baholash metodlari : O‘quvchilarning bilimlarini aniqlash uchun test topshiriqlari, yozma ishlar va og‘zaki savollar asosida tahlil qilish. Testlarda turli darajadagi masalalarni (oddiydan murakkabgacha) qo‘llash. Qiyinchiliklarni aniqlash va tahlil qilish : O‘quvchilarni qiyinlashtirgan mavzular yoki masalalarni ajratib ko‘rsatish. Masalan, ayrim o‘quvchilar ko‘p hadli tenglamalarni faktorlarga ajratishda qiynalishi mumkin. Individual yondashuv : Turli darajadagi bilimga ega bo‘lgan o‘quvchilar uchun moslashuvchan o‘quv strategiyalari ishlab chiqish. Kuchli o‘quvchilar uchun murakkab masalalar, zaifroq o‘quvchilar uchun sodda misollar tanlash. Mavzuni real hayotda qo‘llash imkoniyatlari Amaliy masalalarni yechish : Ko‘p hadli tenglamalar yordamida real hayotiy vaziyatlarda muammolarni hal qilish. Masalan, iqtisodiyot, fizika yoki kimyo sohalaridagi vazifalarni yechishda qo‘llash. Ijtimoiy va texnik ko‘nikmalarni rivojlantirish : Bu bilimlarni turli kasblar (muhandislik, iqtisodiyot, statistika) va kundalik hayotdagi masalalarni hal qilishda qo‘llash. Hayot bilan bog‘liq dars mavzulari : Misol uchun, transport vositalarining harakatini modellashtirish yoki 35 ishlab chiqarish xarajatlarini tahlil qilishda ko‘p hadli tenglamalarning qo‘llanilishi. Dars samaradorligini oshirish yo‘llari Motivatsiya va qiziqish uyg‘otish : Dars jarayonida qiziqarli masalalar va interfaol o‘yinlar qo‘llash. Masalan, guruhli musobaqalar yoki qiziqarli savol-javob sessiyalari orqali mavzuni mustahkamlash. Texnologiyalardan foydalanish : Darsda zamonaviy o‘quv dasturlari (GeoGebra, Desmos) yoki grafik kalkulyatorlardan foydalanish. Bu o‘quvchilarga ko‘p hadli tenglamalarning yechimlarini vizual tarzda tushuntirish imkonini beradi. O‘qituvchi va o‘quvchi o‘rtasidagi muloqot : Dars davomida faol va ochiq muloqotni tashkil qilish, o‘quvchilarning fikrlarini tinglash va ularga ko‘mak berish. Yangi pedagogik texnologiyalar : Differensial yondashuv, aks ettiruvchi ta’lim (reflektiv metod), loyiha asosida o‘rgatish kabi texnologiyalarni tatbiq etish orqali o‘quvchilarni faolroq jalb qilish. Natijalarni umumlashtirish O‘quvchilar ko‘nikmalarini tahlil qilish : O‘quvchilarning masalalarni yechishdagi xatolarini aniqlab, ularga tuzatish bo‘yicha individual maslahat berish. Dars samaradorligi bo‘yicha statistik ma’lumotlar : Sinfdagi o‘quvchilarning mavzuni o‘zlashtirish darajasini jadval yoki grafiklar yordamida tahlil qilish. Tavsiyalar ishlab chiqish : Mavzuni yanada samarali o‘qitish uchun o‘qituvchilarga metodik ko‘rsatmalar va amaliy maslahatlar taqdim etish. 36 FOYDALANILGAN ADABIYOTLAR RO YXATIʻ Prezident asarlari Mirziyoev Sh. M. O zbekiston Respublikasi Prezidentining Oliy Majlisga ʻ Murojaatnomasi. 28.12.2018//O zbekiston Respublikasi Prezidentining rasmiy ʻ veb-sayti. Ilmiy asarlar, Monografiyalar, risolalar 1.R.N. Nazarov, B.T. Toshpo’latov, A.D. Do’sumbetov“Algebra va sonlar nazariyasi” 2-qism.(1995) 2.. T.Sh. Shodiyev “Analitik geometriya va chiziqli algebra ”. 3.A. Hojiyev, A. Faynleyb “Algebra va sonlar nazariyasi”. 4. Iskandarov “Algebra va sonlar nazariyasi”2-qism. 5. D. I. Yunusova , A. S. Yunusov “Algebra va sonlar nazariyasi”. Toshkent. 2009. 317 bet. 6.SH. A. Ayupov, B. A. Omirov, A. X. Xudoyberdiyev, F. H. Haydarov “Algebra va sonlar nazariyasi” . Toshkent. 2019. 319 bet Ilmiy maqolalar 1. Гелфанд И . М . Лекции по линейной алгебре . http://www.mcmee.ru, http://lib.mexmat . ru. 2. Курош А.Г. Курс высшей алгебре http://www.mcmee.ru, http://lib.mexmat . ru Internet saytlari 1. www.arxiv.uz 2. w w w . e d u . u z 37 3. w w w . z i y o n e t . u z 38

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