M A S H Q L A R
55.
Kasrlarni uzluksiz kasrlarga yoying:
.
113
355
)
;
170
99
)
;
33102
103993
)
;
71828
,
2
)
d
c
b
a
56.
Kasrlarni uzluksiz kasrlarga yoying:
.
33102
103993
)
;
100
333
)
;
187
77
)
;
74
247
)
d
c
b
a
57.
Uzluksiz kasrlarga yoyilmasidan foydalanib kasrlarni qiqartiring:
3107
3653
)
;
32671
10027
)
;
2147
6821
)
;
1241
6059
)
;
871
3953
)
e
d
c
b
a
58.
Berilgan kasrni uzluksiz kasrga yoying va uni
4
4
Q
P
kasr bilan almashtiring.
Almashtirish xatosini toping va xatosi ko’rsatilgan holda taqribiy almashtirishga mos
tengligini yozing:
19
359
571
)
;
385
648
)
;
37
29
)
c
b
a
.
59.
Ko’rsatilgan chekli uzluksiz kasrlarga mos oddiy qisqarmaydigan kasrlarni
toping:
(
)
(
)
(
)
(
)
(
)
(
)
(
)
.
7
,
1
,
1
,
2
,
2
,
2
,
13
,
0
)
;
2
,
4
,
5
,
1
,
3
,
2
)
;
5
,
4
,
3
,
2
,
1
,
0
)
;
5
,
2
,
6
,
1
,
2
,
1
,
1
,
2
)
;
5
,
1
,
1
,
1
,
3
,
4
,
2
,
3
,
1
)
;
4
,
3
,
2
,
1
,
1
)
;
4
,
1
,
3
,
2
)
=
−
=
=
=
=
=
=
b
a
g
b
a
f
b
a
e
b
a
d
b
a
c
b
a
b
b
a
a
60.
Tenglamani yeching:
(
)
(
)
7
19
,
2
,
1
,
2
)
;
30
73
4
,
3
,
2
,
)
=
=
x
b
x
a
.
61.
Diofant tenglamalarini yeching:
a
) 41x + 114y = 5;
b
) 19x – 15y = 1;
c
) 23x – 17y = 11;
d
) 53x – 47y = 11;
e
) 35x – 18y = 3;
f
) 85x – 71y = 5;
g
) 41x – 11y = 7.
5-§. Sonli funksiyalar
1.
S o n n i n g b u t u n q i s m i
x
sonning
butun qismi
, ya’ni [
x
] qo’sh tengsizlik bilan
[ ]
[ ]
1
+
≤
≤
x
x
x
yoki
[ ]
x
x
x
≤
<
−
1
; yoki
[ ]
1
0
,
≤
≤
+
=
α
α
x
x
tenglik bilan aniqlanadi va
ant’ye funksiya
deyiladi.
Agar
x
1
va
x
2
sonlardan birortasi butun bo’lsa,
[
x
1
+
x
2
] = [
x
1
] + [
x
2
]
o’rinli bo’ladi.
[ ]
=
m
x
m
x
o’rinli bo’ladi.
m
! ko’paytmaning kanonik yoyilmasiga
p
tub son
+
+
+
S
p
m
p
m
p
m
...
2
darajada keladi, bu yerda
S
son
1
+
<
≤
S
S
p
m
p
tengsizlikdan aniqlanadi.
1-m i s o l.
181
90
cos
2
3
π
−
sonning butun qismini toping.
Yechish. a
∈
Z
va
x
kasr son uchun [
a – x
] =
a
+ [-
x
] formula o’rinli. Bu
formulani qo’llab
20
( )
2
1
3
181
90
cos
2
3
181
90
cos
2
3
=
−
+
=
−
+
=
−
π
π
ni hosil qilamiz.
g
2-m i s o l.
+
n
y
x
ni
+
n
y
n
x
yoki
1
+
+
n
y
n
x
ga tengligini isbotlang.
Yechish.
β
α
+
+
+
=
+
n
y
n
x
n
y
x
bo’lib, bu yerda
1
0
,
1
0
<
≤
<
≤
β
α
. Demak,
[
]
β
α
+
+
+
=
+
n
y
n
x
n
y
x
.
2
0
<
+
≤
β
α
bo’lganligi sababli
[
]
β
α
+
0 yoki 1 ga teng bo’ladi.
g
n
dan katta bo’lmagan va
p
1
, p
2
,..., p
k
tub sonlar bilan o’zaro tub bo’lgan sonlar
sonini quyidagi formula bilan hisoblash mumkin:
(
)
[ ]
( )
.
....
1
....
....
....
....
,...,
,
;
2
1
1
2
3
2
1
1
2
1
1
2
1
−
+
+
−
−
−
+
+
+
+
−
−
−
=
−
−
−
k
k
k
k
k
k
k
k
k
p
p
p
n
p
p
p
n
p
p
p
n
p
p
n
p
p
n
p
n
p
n
n
p
p
p
n
B
3-m i s o l. 180 dan katta bo’lsagan va 5, 7, 11 larga bo’linmaydigan sonlar so-
nini toping.
Yechish. n
= 180 va
p
1
= 5,
p
2
= 7,
p
3
= 11 lar uchun
(
)
[ ]
+
⋅
+
⋅
+
−
−
−
=
11
5
180
7
5
180
11
180
7
180
5
180
180
11
,
7
,
5
;
180
B
113
0
2
3
5
16
25
36
180
11
7
5
180
11
7
180
=
−
+
+
+
−
−
−
=
⋅
⋅
−
⋅
+
.
g
4-m i s o l. 2002! son nechta 0 bilan tugaydi.
Yechish.
Misol yechimi 2002! Ning kanoniy yoyilmasiga 5 nechanchi daraja
bilan kirishini aniqlash masalasiga keltiriladi:
.
499
0
3
16
80
400
3125
2002
625
2002
125
2002
25
2002
5
2002
=
+
+
+
+
=
=
+
+
+
+
Demak, 2002! son 499 ta 0 bilan tugaydi.
g
5-m i s o l. (2
m
)!! ning kanonik yoyilmasiga
p
tub son nechanchi darajada kiri-
shini aniqlang.
Yechish.
(2
m
)!! =
m
! 2
m
bo’lganligi sababli
p
= 2 ga teng bo’lsa,
21
∑
=
+
<
≤
+
k
i
k
k
i
m
m
m
1
1
2
2
,
2
.
p
> 2 bo’lsa,
∑
=
+
<
≤
s
i
s
s
i
p
m
p
p
m
1
1
,
ga teng bo’ladi.
g
2. H a q i q i y s o n n i n g k a s r q i s m i
Haqiqiy
x
sonning kasr qismi {
x
} quyidagi formula bilan aniqlanadi: {
x
} =
x
–
[
x
].
6-m i s o l. {-4,35} ni toping.
Yechish.
{-4,35} = –4,35 – (–5) = 0,65.
g
3. N a t u r a l s o n n i n g b o’ l u v ch i l a r
s o n i v a u l a r y i g’ i n d i s i
Ixtiyoriy natural
a
son uchun
τ
(
a
) va
S
(
a
) funksiyalar mos ravishda
a
son-
ning natural bo’luvchilari soni va ularni yig’indisini ifodalaydi. Bu funksiyalar uchun
quyidagi formulalar o’rinli:
( ) (
)(
) (
)
(
)
( )
∏
∏
=
+
+
+
+
=
−
−
=
−
−
−
−
⋅
−
−
=
+
=
+
+
+
=
n
i
i
i
n
n
n
i
i
n
p
p
p
p
p
p
p
p
a
S
a
i
n
1
1
1
2
1
2
1
1
1
1
2
1
,
1
1
1
1
....
1
1
1
1
1
1
.....
1
1
2
1
α
α
α
α
α
α
α
α
τ
bu yerda
−
∏
=
=
=
n
i
i
i
n
n
p
p
p
p
a
1
2
2
1
1
...
α
α
α
α
a
sonning kanonik yoyilmasi.
Bu funksiyalar multiplikativ, ya’ni agar (
a,b
) = 1 lar uchun
τ
(
ab
) =
τ
(
a
)
τ
(
b
) va S (
ab
) = S(
a
)S(
b
)
o’rinli.
7-m i s o l. 2002 sonni bo’luvchilar soni va ularni yig’indisini toping.
Yechish.
2002 = 2
⋅
7
⋅
11
⋅
13, bundan
(
) ( )( )( )( )
,
16
1
1
1
1
1
1
1
1
2002
=
+
+
+
+
=
τ
(
)
4032
14
12
8
3
1
13
1
13
1
11
1
11
1
7
1
7
1
2
1
2
2002
1
1
1
1
1
1
1
1
=
⋅
⋅
⋅
=
−
−
⋅
−
−
⋅
−
−
⋅
−
−
=
+
+
+
+
S
.
g
8-m i s o l. 2002 sonni barcha bo’luvchilarini toping.
Yechish.
2002=2
⋅
7
⋅
11
⋅
13 – kanonik yoyilmasidan foydalanamiz:
(1+2)(1+7)(1+11)(1+13)=1+2+7+11+13+14+22+26+77+91+143+154+182+286+100
1+2001 – 2002 ning barcha bo’luvchilari yig’indisi va demak har bir qo’shiluvchi iz-
lanayotgan bo’linmalarni beradi.
g
22
9-m i s o l. Natural
a
sonning barcha natural bo’luvchilarining ko’paytmasi
funksiyasi
δ
(
a
) bo’lsa,
( )
( )
a
a
a
τ
δ
=
tenglik to’g’riligini isbotlang.
Yechish. d
1
, d
2
,..., d
τ
(a)
–
a
sonning barcha natural bo’luvchilari bo’lsin. U holda
( )
( )
( )
∏
=
=
=
a
i
а
i
d
d
d
d
a
τ
τ
δ
1
2
1
...
.
( )
−
a
d
a
d
a
d
a
τ
,...,
,
2
1
sonlar
a
ning bo’luvchilaridir, bundan
( )
( )
( )
( )
.
1
1
1
∏
∏
=
=
=
=
a
i
i
a
i
a
i
d
a
d
a
a
τ
τ
τ
δ
δ
(
a
)
uchun hosil bo’lgan tengliklarni ko’paytirib
( )
( )
a
a
a
τ
δ
=
2
ni hosil qilamiz,
bundan
( )
( )
a
a
a
τ
δ
=
.
g
10-m i s o l. 2002 sonining barcha natural bo’luvchilari ko’paytmasini toping.
Yechish.
(
)
8
16
2002
2002
2002
=
=
δ
.
g
11-m i s o l. Barcha natural bo’luvchilari ko’paytmasi 5832 ga teng bo’lgan
natural sonni toping.
Yechish.
( )
6
3
3
2
5832
⋅
=
=
a
a
τ
,
bundan
y
x
a
3
2
⋅
=
va
(
)(
)
(
)(
)
=
+
+
=
+
+
.
12
1
1
6
1
1
y
x
y
y
x
x
Bu sistemaning yechimi:
x
= 1,
y
= 2. Demak,
a
= 18.
g
12-m i s o l. 3 va 4 ga bo’linadigan va 14 ta bo’luvchiga ega bo’lgan sonni
toping.
Yechish.
Misol shartiga ko’ra,
τ
(a) = 14 = 2
⋅
7 == (1+1)(6+1),
demak,
,
2
2
1
1
α
α
p
p
a
=
ya’ni
2
1
3
2
а
a
⋅
=
α
, bu yerda
.
1
,
2
2
1
≥
≥
α
α
Demak,
a
= 2
6
⋅
3 = 192.
g
3. B e r i l g a n m u s b a t s o n d a n k a t t a
b o’ l m a g a n t u b s o n l a r s o n i
π
(
x
) barcha natural
x
lar uchun aniqlangan bo’lib, natural sonlar qatorida
x
dan
katta bo’lmagan tub sonlar sonni bildiradi.
π
(
x
) ni qiymatini tub sonlar jadvali yor-
damida aniqlanadi yoki yetarlicha katta
x
lar uchun taqribiy hisoblash mumkin:
( )
nx
x
x
l
≈
π
va
( )
∫
≈
x
u
n
du
x
2
l
π
.
13-m i s o l.
( )
x
n
x
x
l
=
π
formula yordamida
π
(1000) ni qiymatini toping va
natijaning nisbiy xatosini hisoblang.
Resheniye.
(
)
.
145
9078
,
6
1000
10
3
1000
1000
≈
≈
≈
n
l
π
23
Tub sonlar jadvalidan
π
(1000)=168, demak nisbiy xato
(
)
(
)
%
14
168
145
168
1000
1000
≈
−
=
∆
π
π
.
g
4. E y l ye r f u n k s i ya s i
ϕ
(a)
–
Eyler funksiyasi a
sonning barcha natural qiymatlarida aniqlangan
bo’lib,
a
dan katta bo’lmagan va u bilan o’zaro tub bo’lgan sonlar sonini bildiradi.
ϕ
(1) = 1 deb qabul qilingan. Eyler funksiyasi:
( )
(
)(
)
(
)
1
1
2
2
2
2
1
1
1
1
1
2
1
....
1
1
....
1
1
1
1
−
−
−
−
−
−
=
−
−
−
=
n
n
n
n
n
p
p
p
p
p
p
p
p
p
a
a
α
α
α
α
α
α
ϕ
formula yordamida hisoblanadi, bu yerda
−
=
n
n
p
p
p
a
α
α
α
...
2
2
1
1
sonning kanonik yoy-
ilmasi.
Xususan,
( )
( )
.
1
,
1
−
=
−
=
−
p
p
p
p
p
ϕ
ϕ
α
α
α
Eyler funksiyasi multiplikativ, ya’ni o’zaro tub
a,b,…
,
l
sonlar uchun
(
) ( ) ( ) ( )
l
l
ϕ
ϕ
ϕ
ϕ
...
...
b
a
ab
=
shart bajariladi.
14-m i s o l.
ϕ
(1956) ni hisoblang.
Yechish.
1956=2
2
⋅
3
⋅
163 bo’lganligi sababli
(
)
(
)
(
)(
)
648
162
2
2
1
163
1
3
2
2
1956
2
=
⋅
⋅
=
−
−
−
=
ϕ
.
g
15-m i s o l.
ϕ
(12
⋅
5
⋅
1956) ni hisoblang.
Yechish.
O’zaro tub ko’paytuvchilarni aniqlash uchun ko’paytmani kanonik
yoyilmasini topamiz:
.
163
5
3
2
163
3
2
5
3
2
1956
5
12
2
4
2
2
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
Bundan
(
)
(
) (
)(
)
(
)(
)
=
−
−
−
−
=
⋅
⋅
⋅
=
⋅
⋅
1
163
1
5
3
3
2
2
163
5
3
2
1956
5
12
2
3
4
2
4
ϕ
ϕ
.
31104
162
4
6
8
=
⋅
⋅
⋅
=
g
16-m i s o l.
(
)
600
5
3
=
⋅
y
x
ϕ
tenglamani yeching.
Yechish.
600 = 2
3
⋅
3
⋅
5
2
dan
(
)
.
5
3
2
5
3
2
3
⋅
⋅
=
⋅
y
x
ϕ
Boshqa tomondan
( ) (
)(
)
1
1
5
5
3
3
5
3
−
−
−
−
=
y
y
x
x
y
x
ϕ
.
Demak,
2
3
1
1
5
3
2
4
5
2
3
⋅
⋅
=
⋅
⋅
⋅
−
−
y
x
yoki
2
1
1
5
3
5
3
⋅
=
−
−
y
x
va
x
= 2,
y
= 3.
g
17-m i s o l.
a
= 72 uchun
Gauss formulasini
to’g’riligini ko’rsating:
( )
∑
=
a
d
a
d
/
ϕ
.
Yechish.
Gauss formulasi
( )
∑
=
a
d
a
d
/
ϕ
da
a
= 72 deb olamiz:
a
= 72 = 2
3
⋅
3
2
.
72 ning barcha bo’luvchilari:
(1 + 2 + 2
2
+ 2
3
)(1 + 3 + 3
2
).
( ) ( ) ( )
( ) ( )
[
]
( ) ( )
( )
[
]
=
+
+
+
+
+
∑
=
2
3
2
72
/
3
3
1
2
2
2
1
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
ϕ
d
d
=
(
) (
)
.
72
9
8
6
2
1
4
2
1
1
a
=
=
⋅
=
+
+
+
+
+
g
24
18-m i s o l.
ϕ
(
x
) =
p
– 1 tenglamani yeching.
Yechish. x
=
r
α
⋅
y
deb olamiz, bu yerda (
y, r
) = 1.
r
α
- 1
ϕ
(
y
) = 1, bundan
α
= 1 va
ϕ
(
y
) = 1. Demak,
r
= 2 da tenglama yagona
x
= 2
(chunki bu holda
y
= 1);
r
> 2 da tenglama ikkita:
x
=
r
; 2
r
yechimga ega.
g
19-m i s o l. Eyler funksiyasining xossalaridan foydalanib tub sonlar soni chek-
siz ko’pligini isbotlang.
Yechish. r
1
, r
2
,…,r
k
– barcha tub sonlar bo’lsin, u holda
a = r
1
⋅
r
2
…r
k
son uchun
ϕ
(a) = (r
1
– 1) (r
2
– 1)…(r
k
– 1)
bo’ladi. Boshqa tomondan
ϕ
(a)=
1, chunki ixtiyoriy birdan farqli va
a
dan katta bo’lmagan son oddiy
bo’luvchiga ega va bu bo’luvchi
r
i
lardan birortasiga teng, shu sababli bu son
a
bilan
o’zaro tub bo’la olmaydi. Demak,
(r
1
–
1
)(r
2
–
1
)…(r
k
–
1
)=
1, lekin bu tenglik
k
= 2 dan
boshlab o’rinli emas, (2-1)(3-1) > 1 hosil qilingan qarama-qarshilik tub sonlar soni
cheksizligini bildiradi.
5. M y o b i u s f u n k s i ya s i
Barcha natural sonlar uchun aniqlangan
( ) ( )
≠
≠
=
−
=
=
булинса
квадратига
сон
туб
a
агар
j
i
р
р
р
р
р
a
агар
a
агар
a
j
i
k
k
,
0
,
...
,
1
1
,
1
2
1
µ
ko’rinishdagi funksiyaga
Myobius funksiyasi
deb ataladi.
Bu funksiya multiplikativdir, ya’ni agar (
a,b
)=1 bo’lsa,
)
(
)
(
)
,
(
b
a
b
a
µ
µ
µ
=
.
Agar
θ
(a)
– ixtiyoriy multiplikativ funksiya bo’lsa, u holda
( ) ( )
( )
(
)
∑
≠
∏ −
=
=
a
d
a
p
a
агар
р
a
агар
d
d
/
/
.
1
,
1
1
,
1
θ
θ
µ
Agar bu formulada
( )
1
≡
a
θ
va
( )
a
a
1
=
θ
deb olsak quyidagshi formulalarni ho-
sil qilamiz:
( )
( )
∑
>
∏
−
=
=
∑
≠
=
=
a
d
a
p
a
d
a
агар
p
a
агар
d
d
a
агар
a
агар
d
|
/
|
.
1
,
1
1
1
,
1
1
,
0
1
,
1
µ
µ
Agar butun
a
lar uchun
f (a)
– funksiya birqiymatli bo’lib,
25
( )
( ) (
)
∑
>
=
a
d
d
d
f
a
F
|
0
o’rinli bo’lsa, u holda
( )
( )
∑
=
a
d
d
a
F
d
a
f
|
µ
tenglik o’rinlidir (
Myobiusning teskarilash formulasi
).
20-m i s o l.
µ
(2002) ni hisoblang.
Yechish.
2002 = 2
⋅
7
⋅
11
⋅
13 dan
µ
(2002) = (-1)
4
= 1 kelib chiqadi.
g
21-m i s o l.
( )
∑
=
=
a
d
а
d
|
18
0
µ
uchun to’g’riligini isbotlang.
Yechish.
18 ning bo’luvchilari: 1, 2, 3, 6, 9, 18. Bundan
( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
0
0
0
1
1
1
1
18
9
6
3
2
1
)
(
18
|
=
+
+
+
−
+
−
+
=
=
+
+
+
+
+
=
∑
d
d
µ
µ
µ
µ
µ
µ
µ
.
22-m i s o l.
( )
∏
−
=
∑
a
p
a
d
p
d
d
|
|
1
1
µ
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