Bir noma'lumli ko‘p hadli tenglamalar - Algebra - Kurs ishlari

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Bo'lim	Kurs ishlari
Fan	Algebra

Sotuvchi

Mohigul Xolyigitova

Ro'yxatga olish sanasi 25 Oktyabr 2024

7 Sotish

Bir noma'lumli ko‘p hadli tenglamalar

Sotib olish

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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