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; d)
32 10 2 15 4 6 6 + + +
s co x s co x s co ;
e) ( ) x in s x in s x in s x in s 2 6 4 2 6 2 8 128 1 − − + ; f) ) ( ( ) ( ) ( ) [ ] inx s x s co x in s x s co x in s x s co x in s x s co + + − + + + − 35 3 3 21 5 5 7 7 7 64 1 .
4-§.
46. κ κ − = n n n C C 2 2 2 ekanligini hisobga olinsa, a) va b) tengliklar 1-misolga keltiriladi. 47. a) va b) tenglamalarni 2-misoldagidek hosil qilish mumkin, bunda: ( ) ( )
n n 2 2 1 1 2 ε ε ε ε + + + + , ( ) ( ) n n n 2 2 1 1 2 ε ε ε ε + + + + . 48. 4-misolga qarang. 49. 5-misoldagi (*) ayniyatdan k x = bo’lganda kelib chiqadi. 50. Yechish. Chap tomondagi ifoda quyidagi ko’phaddagi n x oldidagi koeffisiyentan iborat: ( )
) ( ) ( ) ( ) = + − + + + + + − + − − − n k n k n n n n n n x x x x x x x x 1 1 1 1 1 2 1 K
( )
( ) ( ) ( )
( )
n k n n k n k n n n x x x x x x x − + − − − − + + = − + + − + = 1 1 1 1 1 1 1 K . Oxirgi ifodada n x oldidagi koeffisiyent ( )
1 1 − − ga tengligi ravshan. 51. 7-misoldan kelib chiqadi. 52. a) Yechish. ( ) (
) ( ) m m m x x x 2 1 1 1 − = − + ko’paytmani qaraymiz. natijada, ( )
( ) ∑ ∑ ∑ = = = − = − m t m k k k m k t m s s s m s x C x C x C 0 0 2 0 1 1 , shuning uchun 38 ( ) ( ) k m k t m k t s s m s C C C 1 1 2 − = − ∑ = +
Avvalo faraz qilaylik, m – juft, ya’ni n m 2 = , n k = bo’lsin. U holda ( ) ( )
n n n nt n m n t s s n s C C C 2 2 2 2 2 1 1 − = − − = + ∑ . Bu yerdan ( )
( ) ( )
n n n n s s n s C C 2 2 0 2 2 1 1 − = − ∑ = ni hosil qilamiz; b) agar m – toq bo’lsa, 1 2
= n m deb olamiz.
( ) ( ) ( ) m m m x x x 2 1 1 1 − = − + tenglikning chap tomonidagi 1 2 + n x oldidagi koeffisiyent ( ) ( )
( ) ∑ ∑ + = + + = + + + − = − 1 2 0 2 1 2 1 2 1 2 1 2 1 1 n s s n S n t s t n s n s C C C ga teng. Lekin qaralayotgan tenglikning o’ng tomonidan ko’rinadiki, bu koeffisiyent nolga teng bo’lishi kerak (chunki yoyilmasida x ning toq darajali hadlari qatnashmaydi). Shuning uchun ( ) ( )
0 1 1 2 0 2 1 2 = − ∑ + = +
s s n s C va tenglik isbot bo’ldi. 53. a) 9-misolda ϕ ni
ϕ π − 2 ga almashtiring; b) va s) lar ham 9-misol va a) ga o’xshash keltirib chiqariladi. 54. 10-misolga o’xshash. 55. a) (
2 2 2 2 x n s co x s co n n + ; b) ( ) 2 2 2 2 x n in s x s co n n + . 56. x in s nx in s n 2 4 4 2 − . Ko’rsatma: 2 1 2 α α s2 co in s − = formuladan foydalaning. 58. a)
( ) ( ) 2 4 1 1 1 2 x in s x n s nco nx s co n − + − + ; b) ( ) ( ) 2 4 1 1 2 x in s x n in s n innx s n + − + .
1 2
2 1 − + + + + n na K α α
ko’rinishdagi yig’indini hisoblash uchun uni α − 1 ga ko’paytirish foydali.
5-§. 59. a)
1 ± ; b) 2 3 2 1 , 1 i ± − ; c) i ± ± , 1 ; d) ( )
i i ± ± ± ± 1 2 2
,
, 1 ;
e) 2 2 3 2 3 2 1
,
, 1 i , i i ± ± ± ± ± ± ;
g) ( )
4 2 6 4 2 6 4 2 6 4 2 6 2 2 3 1 2 3 2 3 2 1 1 + ± − ± − ± + ± ± ± ± ± ± ± ± ± i ; i ; i ; i , i i, ,
60. a) -1; b) 2 3 2 1 i ± − ; c) i ± ; d) ( ) i ± ± 1 2 2 ; e) 2 2 3 i ± ± ; 39 f) 4 2 6 4 2 6 4 2 6 4 2 6 + ± − ± − ± + ±
; i .
61. a) 16 2 16 2
in s i k s co k π π ε + = belgilashni kiritib, quyidagilarni hosil qilamiz: 1 ko’rsatkichga 0 ε
2 ko’rsatkichga 8 ε tegishli; 4 ko’rsatkichga 12 4
ε ε tegishli; 8 ko’rsatkichga 14 10 6 2 , , , ε ε ε ε tegishli; 16-darajali boshlang’ich ildizlar , 15
13 , 11 , 9 , 7 5 3 1 , , , ε ε ε ε ε ε ε ε b)
20 2 20 2 , πκ πκ ε κ in s i s co + = belgilashni kiritib, quyidagilarni hosil qilamiz: 1 ko’rsatkichga 0 ε
2 ko’rsatkichga 10 ε tegishli; 4 ko’rsatkichga 15 ,
ε ε tegishli; 5 ko’rsatkichga 16 12 8 , 4 , , ε ε ε ε tegishli; 10 ko’rsatkichga 18 ,
, 6 , 2 ε ε ε ε tegishli; 20-darajali boshlang’iya ildizlar 19 , 17 , 13 , 11 , 9 , 7 , 5 , 3 , 1 ε ε ε ε ε ε ε ε ε s)
24 2 24 2 , πκ πκ ε κ in s i s co + = belgilashlarni kiritib, quyidagilarni hosil qilamiz: 1 ko’rsatkichga 0 ε
2 ko’rsatkichga 12 ε tegishli; 3 ko’rsatkichga 16 ,
ε ε tegishli; 4 ko’rsatkichga 18 , 6 ε ε tegishli; 6 ko’rsatikichga 20 ,
ε ε tegishli; 8 ko’rsatkichga 21 , 15 , 9 , 3 ε ε ε ε tegishli; 12 ko’rsatkichga 22 ,
, 10 , 2 ε ε ε ε tegishli 24-darajali boshlang’ich ildizlar 23 , 19 , 17 , 13 , 11 , 7 , 5 , 1 ε ε ε ε ε ε ε ε . 62.
ε − 1 2 .
63. 0, agar 1 > n bo’lsa. 64. ε
− 1
, agar 1
ε bo’lsa; ( )
1 +
n , agar
1 = ε bo’lsa. 65.
( ) ( ) 2 2 1 2 1 ε ε − + − − n n , agar
1 ≠ ε bo’lsa; ( )( ) 6 1 2 1 + + n n n , agar
1 = ε bo’lsa. 40 66. a) 2 n − ; b) n ctg n π 2 − ;
67. a) 1; b) 0; c) -1. 68. Yechilishi: Agar z berilgan tenglamani qanoatlantirsa, u holda n b z a z λ µ = − − bo’ladi. Berilgan ikki nuqtalargacha bo’lgan masofalar nisbati o’zgarmas bo’lgan nuqtalar to’plami aylanadan iborat (xususiy holda, µ λ = bo’lsa, bu to’plam to’g’ri chiziq bo’ladi). 69. a) (
1 1 0 2
, , k n k ictg K = − π ; b) ( ) 1 1
5 ,n- , k n k ctg x K = = π ; c) ( ) 1 1 0 4 3 4 3
, , k n k ctg x K = + = π . Ko’rsatma. ( ) ( ) 1 , , 1 , 0
2 3 4 2 3 4 , 3 3 − = + + + = − = − = + n k n k in s i n k s co i i x i x n k k K π π α α tenglamani qarang; d) (
1 1 0 2 2 ,n- , , k n k actg K = + ϕ π . 70. Yechish. ϕ ϕ
s i s co A + = bo’lsin. U holda 2 1 1 k ix ix η = − + , bu yerda ( ) 1 1 0
2 2 2 2 ,m- , , k m k sin i m k s co k K = + + + = π ϕ π ϕ η . Bundan
( ) (
) m k tg i i x k k k k k k 2 2 1 1 1 1 2 2 π ϕ η η η η η η + = + − = + − = − − . 71. Yechish. − ε
1 −
х va
1 −
х larning umumiy ildizi; s - ε ildiz tegishli bo’lgan ko’rsatkich bo’lsin. U holda s - a va v ning umumiy bo’luvchisi bo’ladi shuning uchun faqat s=1 va ε =1 bo’lishi mumkin. Teskarisi ko’rinib turibdi. 72. α va β - 1 ning a va b -darajali boshlang’ich ildizlari bo’lsin. ( ) 1
s αβ bo’lsin. U holda 1 = bs α ; 1 =
β . Demak
bs a ga bo’linadi, as b ga bo’linadi. Natijada s ab ga
bo’linadi. λ - 1 ning ab -darajali boshlang’ich ildizi bo’lsin. U holda s β α λ κ = (9-misolni qarang). κ α
a a < 1 ko’rsatkichga tegishli bo’lsin. U holda ( ) ( ) 1 2 1 1 = = b a s b a b a a β λ κ , bu esa mumkin emas. Xuddi shunday, S β - 1 ning b- darajali boshlang’ich ildizi bo’lishini ko’rsatish mumkin. 73. 72-masaladan kelib chiqadi. 74. Yechish. Avvalo r α dan oshmaydigan barcha r ga karrali sonlarni yozib olamiz. Bular 1 ⋅
⋅
α -1 r. Bunday sonlar 1 − α p ta. Natijada, Eyler funksiyasining ta’rifiga ko’ra, ( )
− = − = −
р р р р 1 1 1 α α α α ϕ . U holda 73-masalaga ko’ra
( ) ( ) ( ) ( )
− − − = = κ ακ α α ϕ ϕ ϕ ϕ p p р n p p p n n 1 1 ... 1 1 1 1 ... 2 1 2 1 1 1 . 41 75. Yechish. Agar ε - birning n -darajali boshlang’ich ildizi bo’lsa, u holda uning qo’shmasi ε ham 1 ning n -darajali boshlang’ich ildizi bo’ladi. Bunda ε ≠± 1, chunki 2 >
. 76. a) ( ) 1 1 − =
x X ; b)
( ) 1 2 + =
x X ; c)
( ) 1 2 3 + + = x x x X ; d) ( ) 1 2 4 + = x x X ; e)
( ) 1 2 3 4 5 + + + + =
x х х x X ; f)
( ) 1 2 6 + − = х x x X ; g)
( ) 1 2 3 4 5 6 7 + + + + + + = х x х х х х x X ;
h) ( )
1 4 8 + =
x X ; i)
( ) 1 3 6 9 + + =
х x X ;
j) ( )
1 2 3 4 10 + − + − = х x х х x X ;
k) ( )
1 2 3 4 5 6 7 8 9 10 11 + + + + + + + + + + = x x х х х х х х х х x X ;
l)
( ) 1 2 4 12 + − =
х x X ; m)
( ) 1 3 4 5 7 8 15 + − + − + − = x х х х х х x X ;
( ) 1 2 2 ) 2 5 6 7 8 9 12 13 14 15 16 17 20 22 24 26 28 31 32 33 34 35 36 39 40 41 42 43 46 47 48 105 + + + − − − − − + + + + + + − − − − − − + + + + + + − − − − + − + =
х х х x х х х х х х x х х х х х х x х х х х х х x х х х х х х x X n
77. ( )
1 ...
2 1 + + + + = − − х x x x X р р р .
78. ( )
( ) ( ) 1 ... 1 1 2 1 1 + + + + = − − − − − m p m p р m р р m р х x x x X . Ko’rsatma. 1 1
− m р x
ning barcha ildizlari va faqat ular 1 −
р x ning boshlang’ich ildizlari bo’ladi. 79. Yechish. ( )
n ϕ α α α ,... , 2 1 - 1 ning n-darajali boshlang’ich ildizlari bo’lsin. U holda (72-masalaga qarang) ( )
α − , ( ) ( ) ( )
ϕ α
− − ,..., 2 sonlar 1 ning 2n – darajali boshlang’ich ildizlari bo’ladi. ( ) (
) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 ... α α α α ϕ ϕ − − − = + + + =
x x x x X n n n ( ) ( ) ( ) т х х ϕ α α − − − − ... 2 , yoki (75-masalaga qarang) ( ) ( )
x X x X n n − = 2 .
80. Yechish. nd in s i nd s co πκ πκ ε κ 2 2 + = - 1 ning nd-darajali boshlang’ich ildizi bo’lsin, ya’ni k va n o’zaro tub sonlar. k ni n ga bo’lib, k = nq+r, 0 yerdan:
n r q in s i d n r q s co π π π π ε κ 2 2 2 2 + + + = , ya’ni
κ ε - d darajali ildizning qiymatlaridan biri bo’ladi va n r in s i n r s co r π π η 2 2 + = ; r η - 1 ning n darajali boshlang’ich ildizi bo’ladi, chunki r va n ning har bir umumiy bo’luvchisi k va n ning umumiy bo’luvchisi bo’ladi. n r in s i n r s co r π π η 2 2 + = - 1 ning n-darajali boshlang’ich ildizi bo’lsin, ya’ni r va n o’zaro tub sonlar. Quyidagi sonlarni qaraymiz 42 ( ) ( ) nd nq r in s i nd nq r s co d n r q in s i d n r q s co q + + + = + + + = π π π π π π ε 2 2 2 2 2 2 , bu yerda ; 1 ..., 2 , 1 , 0 − = d q
ε - 1 ning nd darajali boshlang’ich ildizi bo’ladi. Haqiqatan, agar nq r + va nd sonlar bir vaqtda r tub songa bo’linsa, n va r sonlar ham p ga bo’linar edi. Bu esa mumkin emas. 81. Yechish. ( )
′ ϕ ε ε ε ,... , 2 1 - 1 ning n ′
darajali ildizlari bo’lsin. U holda − Π = = κ ϕ κ ε // / 1 // /
n n n x x Х . ( )( ) ( ) // , 2 , 1 , ....
n x x x κ κ κ ε ε ε − − − -
− κ ε //
x ning
chiziqli ko’paytuvchilarga yoyilmasi bo’lsin. U holda ( ) i n i n i n n x x Х ,. // / 1 1 // / κ ϕ κ κ ε − Π = = = = = . 80- masalaga ko’ra har bir i x ,. κ ε − chiziqli ko’paytuvchi ( ) x X n yoyilmaga kiradi va aksincha. Bundan tashqari, ( )
( ) / // n n n ϕ ϕ = bo’lganligi uchun ( )
va
// / n n x X larning darajalari teng. 82. Ko’rsatma. 77, 78, 72- masalalardan foydalaning va 1) r – tub bo’lsa, ( )
1 − = р µ ,; 2) r – tub, α> 1 bo’lsa, ( ) 0 = α µ
; 3) a va b o’zaro tub bo’lsa, ( ) ( ) ( ) b а аb µ µ µ = . 83. Yechish. 1 ning barcha n-darajali ildizlari yig’indisi 0 ga teng. 1 ning har bir n- darajali ildizi n ning bo’luvchisi bo’lgan d ko’rsatkichga tegishli i obratno, to ( ) ∑
n d d 0 µ . 84. Yechish. n in s i n s co πκ πκ ε κ 2 2 + = ildiz n 1 ko’rsatkichga tegishli bo’lsin. U holda x- ε κ ko’paytuvchi faqat shunday x d -1 ikki hollarda qatnashadiki, d son n 1 ga bo’linadi. Bunda d n ning n
karrali barcha bo’luvchilari to’plamida, d n esa
1 n n ning barcha bo’luvchilari to’plamida o’zgaradi. Shunday qilib, x- ε κ ko’paytuvchi o’ng tomonda ( ) ∑
1 1
n d d µ ko’rsatkich bilan qatnashadi. Agar 1 1 ≠ n n bo’lsa, bu yig’indi 0 ga, n = n 1
85. Yechish. Agar α
n = , r – tub son bo’lsa, ( ) p X n = 1 bo’ladi. Agar κ α κ α α p p p n ...
2 2 1 1 = ( κ p p p ,....,
, 2 1 – har xil tub sonlar) bo’lsa, u holda (81-masalaga qarang)
( ) ( )
1 1
n n X X = ; bunda n / = κ p p p ...
2 1 . 43 Endi n= κ p p p ...
2 1 ; κ κ
n n = ≥ 1
; 2 bo’lsin. 1 n ning barcha bo’luvchilarini hosil qilish uchun n ning barcha bo’luvchilariga ularning r k ga ko’paytmalarini qo’shish yetarli. Shuning uchun ( )
( ) ( ) ( ) ( ) [ ]
( ) κ µ µ µ
n n k dp n k dp n d d n d n d d n d n d n x X x X x x x x X / 1 / 1 / 1 1 1 −
= = − Π ⋅ − Π = − Π = . Bu yerdan ( ) 1
= n X .
86. Yechish. 1) n – birdan katta toq son bo’lsin. U holda (79-masalaga qarang) ( )
( ) 1 1 1 2 = = −
n X X ;
2) n = 2 κ
bo’lsin, u holda 1 1 1 2 2 + = − − =
n n n х х х X va
( ) 1 − n X k=1 bo’lganda 0 ga, k>1 bo’lganda 2 ga teng. 3)
1 2n n = , 1 n - birdan katta toq son bo’lsin. U holda (79-masalaga qarang) ( ) ( )
1 1 1 n n X X = − va natijada ( )
1 −
X
α p n = 1 bo’lganda (r – tub son) p ga, p n ≠ 1 bo’lganda 1 ga teng. 4) 1
n κ = , κ >1, s s p p p n α α α ....
2 2 1 1 1 = ( s р р р ,...
, 2 1 - har xil toq sonlar) bo’lsin. Bu
holda (81-masalaga qarang) ( )
( ) λ
р р р Х х X s n ....
2 1 2 = , bunda 1 1 1 1 1 ... 2 − − − =
s p p α α κ λ . Bu yerdan kelib chiqadiki, ( ) ( )
1 1 1 = = − n n X X . 87. Yechish. ( ) n ϕ ε ε ε , , , 2 1 K - 1 ning boshlang’ich ildizlari bo’lsin: ( ) ( )
( ) [ ] ( ) ( ) 2 2 2 2 2 1 2 1 3 1 2 1 n n n n S ϕ ϕ ϕ ε ε ε µ ε ε ε ε ε ε + + + − = + + + = − K K . 1) m – toq son bo’lsin. Bu holda 2
ε 1 ning n–darajali boshlang’ich ildizi va faqat j i = bo’lganda 2 2
i ε ε = bo’ladi. Shuning uchun ( ) ( )
n n µ ε ε ε ϕ = + + + 2 2 2 2 1 K va
( ) [ ] ( ) 2 2 n n S µ µ − = . 2) 1 1 ; 2 n n n = - toq son bo’lsin. Bu holda i ε (72-masalaga qarang) 1 ning 1 n darajali boshlang’ich ildizi
bo’ladi va
shuning uchun
(1) ga
qarang) ( )
( ) ( )
n n n µ µ ε ε ε ϕ − = = + + + 1 2 2 2 2 1 K . Shunday qilib, bu holda ( ) [ ] ( ) 2 2 n n S µ µ + = . 3)
1 2 n n k = , 1 >
, 1
- toq son bo’lsin. Bu holda 2
ε ildiz
2 n ko’rsatkichga tegishli bo’ladi. 80-masalaga ko’ra ( )
n ϕ ε ε ε K , , 2 1 lar
2 2 1 , , , n ϕ η η η K larning kvadrat ildizlaridan 44 iborat bo’ladi, bu yerda 2 2 1 , , , n ϕ η η η K - 1 ning 2
- darajali boshlang’ich ildizlari. Bu yerdan kelib chiqadiki, ( )
= = + + + = + + + 2 - S ; 2 2 2 2 2 1 2 2 2 2 1 n n n n µ µ η η η ε ε ε ϕ ϕ K K . 88. Yechish. y ning istalgan qiymatlarida ( ) ∑ ∑ ∑ − = + − + = − = = = = 1 0 2 1 2 1 0 2 n S S y n y y x x n x x S ε ε ε ;
n ning toq qiymatlarida ( ) ∑ ∑ ∑ ∑ − = − = + − − = − − = − = = = = ′ 1 0 1 0 2 2 1 0 2 / 1 0 2 ,
y n s s y y n y y n y y e S S S S ε ε ε
∑ ∑ ∑∑ − = − = − = − = + = = = 1 0 1 0 2 2 1 0 1 0 2 2 n s n y ys s n y n s s ys ε ε ε ( )
n n n y y s n s s = + ∑ ∑ − = − = 1 0 2 1 1 2 ε ε . n juft bo’lganda ( )
− + = + = 2 2 2 / 1 1
n n n n SS ε , chunki n ga bo’linmaydigan s 2 uchun 0 1 0 2 = ∑ − =
y sy ε . Shunday qilib, n S = , agar n – toq bo’lsa va ( ) − + = 2 1 1
n S , agar n – juft bo’lsa.
6-§.
89. Yechish. n bi a u n + + = 1 ni trigonometrik shaklga keltiramiz va n n u ning
absolyut qiymati va argumentining limitlarini topamiz. Natijada quyidagini hosil qilamiz:
n n a n n n n n n u g ar e n b a n a u r ϕ = →
+ + + = = ; 2 1 2 2 2 2 , bunda 0
n → = n nr b sin ϕ . , 0 → n ϕ deb hisoblab, b r b n n n n n → ⋅ = ϕ ϕ ϕ sin
ni hosil qilamiz. Shunday qilib, ( )
in is osb c e u m i l a n n n + = ∞ → . 90. Yechish. ( )( ) ( ) ( ) 2 2 2 1 2 2 1 1 ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ + + + = + + in s i s co in s i s co in s i s co
formula ( ) 2 1 2 1 ϕ ϕ ϕ ϕ + = i i i e e e formulaga, ya’ni bir xil asosli darajalarni ko’paytirish qoidasiga aylanadi. Xuddi shunday, Muavr formulasi ( )
n i n i e e ϕ ϕ = formulaga aylanadi. 91. a) ( ) i k π 1 2 1 + + ; b)
( )
k n l π 1 2 2 + + ; c)
( ) 2 1 4
k π + ; 45 d) ( ) 4 1 8
k π + ; e) –1; f) 2 2 n i k e l + − π ; g) 4 2 π π +
e . 92. i k x s arcco i π 2 + .
93. Yechish. ϕ ϕ ϕ ϕ ϕ ϕ ϕ
i i i e e e e i s co in s tg − − + − ⋅ = = 1 ni hosil qilamiz. x tg = ϕ bo’lsin. U holda
π ϕ ϕ k ix ix n l i ix ix e i + − + ⋅ = − + = 1 1 2 1
, 1 1 2 .
Foydalanilgan adabiyotlar
1. B.L. Van der Varden. Algebra. M., Nauka, 1976. 2. Kostrikin A.I. Vvedeniye v algebru. M., 1977, 495 str. 3. Leng S. Algebra. M. Mir, 1968. 4. Faddeyev D.K. Leksii po algebre. M., Nauka, 1984, 415 st. 5. Faddeyev D.K., Sominskiy I.S. Sbornik zadach po vysshey algebre. M., Nauka, 1977. 6. Sbornik zadach po algebre pod redaksiyey. A.I. Kostrikina, M., Nauka, 1985. 7. Xojiyev J., Faynleb A.S. Algebra va sonlar nazariyasi kursi, Toshkent, «O’zbekiston», 2001. 8. Narzullayev U.X., Soleyev A.S. Algebra i teoriya chisel. I-II chast, Samarkand, 2002.
46
Mundarija
1-§. Algebraik shakldagi kompleks sonlar …………………….3 2-§. Kompleks sonning geometrik tasviri va trigonometrik shakli……………………………….……..8 3-§. Darajaga ko’tarish va ildiz chiqarish ……..……………. 15 4-§. Yig’indi va ko’paytmalarni kompleks sonlar yordamida hisoblash ………………………………….. 19 5-§. Birning ildizlari ………………………………………. 26 6-§. Kompleks o’zgaruvchining ko’rsatkichli va logarifmik Funksiyalari …………………………………………….. 31 Javoblar. Ko’rsatmalar. Yechilishlar ………………………………… 32 Foyadalanilgan adabiyotlar ………………………………………46
47
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'alas soloh
Hayya 'alas
mavsum boyicha
yuklab olish