.
55
56.
(
)
2
,
2
,
1
,
2
,
3
74
247
)
=
a
, munosib kasrlari:
;
74
277
;
35
131
,
4
15
,
3
11
,
3
4
,
1
3
(
)
,
3
,
2
,
2
,
0
187
77
)
=
b
munosib kasrlari:
;
12
7
,
5
2
,
2
1
,
1
0
(
)
,
33
,
3
,
3
100
333
)
=
c
munosib kasrlari:
;
100
333
;
3
10
;
1
3
(
)
,
292
,
1
,
15
,
7
,
3
33102
103993
)
=
d
munosib kasrlari:
.
33102
103993
;
113
355
;
106
333
;
7
22
;
1
3
57
.
a
)
37
29
sonni uzluksiz kasrga yoyamiz:
(
)
2
,
1
,
1
,
1
,
3
,
1
,
0
7
29
=
. Sxema yor-
damida
64
k
0 1
2
3
4
5
6
q
k
0 1
3
1
1
1
2
P
k
1 0 1
3
4
7
11
29
Q
k
0 1 1
4
5
9
14
37
kasrlarni topamiz
bilan
i
ortig
Q
Q
Q
Р
'
78
,
0
9
7
;
01
,
0
008
,
0
126
1
14
9
1
1
9
7
5
4
4
4
≈
<
≈
=
⋅
=
=
.
Shunday qilib,
(
)
.
78
,
0
01
,
0
9
7
37
29
=
+
≈
b
a
Q
Р
<
4
4
bo’lganligidan xatoni + ishora bi-
lan olinadi. 7 ni 9 ga bo’lganda bo’linma ortig’i bilan olinishi sababi
9
7
jami bilan
yaqinlashishi bo’lganligidir. O’nli yaqinlashish
78
,
0
37
29
≈
da xato ko’rsatilmaganligi
sababi bu xatoni maxsus hisoblashdadir, ya’ni bu + 0, 008 xato va 7 ni 9 ga
bo’lganda yaxlitlash xatolar yig’indisi.
(
)
6830
,
1
0003
,
0
41
69
385
648
)
≈
+
≈
b
;
(
)
.
5909
,
1
0005
,
0
22
35
359
571
)
≈
−
≈
c
59
.
.
11953
893
)
;
215
159
1
)
;
225
157
)
;
552
1421
)
;
1810
2633
)
;
43
73
)
;
19
43
)
g
f
e
d
c
b
a
−
60
. a)
x
= 2; b)
x
= 2.
61
. a)
x
= - 125 – 114
t
,
y
= 45 + 41
t
;
b)
x
= 4 + 15
t
,
y
= 5 + 19
t
;
c)
x
= 33 + 17
t
,
y
= 44 + 23
t
;
d)
x
= 88 + 47
t
,
y
= 99 + 53
t
;
e)
x
=- 3 + 18
t
,
y
= 6 + 35
t
;
f)
x
=- 25 + 71
t
,
y
=- 30 + 85
t
;
g)
x
= - 28 + 11
t
,
y
=- 105 + 41
t
,
t
∈
Z
.
5-§
62
.
a
) – 3;
b
) 11;
c
) 1;
d
) 2;
e
) 3;
f
) 2;
g
) – 2;
h
) – 2; agar
;
1000
,
1
,
1000
______
=
−
>
abcd
agar
va
abcd
i) 7; j) – 3.
63.
Yechish.
[ ]
[ ]
,
1
0
,
1
0
sin,
'
ва
2
1
2
1
<
≤
<
≤
+
=
+
=
θ
θ
θ
θ
yerda
bu
l
bo
y
y
x
x
u holda
[ ] [ ]
(
)
[
] [ ] [ ]
;
,
1
0
.
2
1
2
1
y
x
y
x
Agar
y
x
y
x
+
=
+
<
+
≤
+
+
+
=
+
θ
θ
θ
θ
agar
65
2
1
2
1
<
+
≤
θ
θ
bo’lsa
[
] [ ] [ ]
y
x
y
x
+
>
+
bo’ladi. Natijalarni birlashtirsak,
[
x + y
]
≥
[
x
] + [
y
] ni hosil qilamiz.
64
. Yechish.
[
x
] ni ta’rifiga ko’ra, masala shartiga asosan
ax
=
m
+
θ
, bu yerda 0
≤
θ
< 1 va
a
≠
0, bu tenglikdan
a
m
x
θ
+
=
ni hosil qilamiz.
65
. Yechish.
12,4
m
= 86 +
θ
, bu yerda 0
≤
θ
< 1. Tenglikni 5 ga
ko’paytiramiz: 62
m
= 430 + 5
θ
, bundan
.
62
5
58
6
62
5
430
θ
θ
+
+
=
+
=
m
0
≤
θ
< 1 dan 0
≤
5
θ
< 5 va
m
butun musbat son bo’lishi uchun
t
=
+
62
5
58
θ
butun bo’lishi lozim.
t
= 1 deb olsak,
7
ва
5
4
=
=
m
θ
ni hosil qilamiz.
66.
Yechish.
.
4
3
4
;
4
1
4
3
4
1
4
−
=
=
−
=
=
+
=
+
=
p
n
p
p
n
p
n
p
n
p
67.
Yechish.
,
1
0
,
,
0
,
1
<
≤
+
=
<
≤
+
=
m
r
m
r
q
m
a
ёки
m
r
mq
a
bu
yerdan
.
m
r
a
m
a
и
m
a
q
−
=
=
68.
.
2
1
2
1
2
1
2
−
=
=
+
=
+
=
m
k
k
m
k
m
69.
a)
Yechish
. 2
≤
x
2
< 3 yoki
2
3
,
3
2
−
≤
<
−
<
≤
x
bundan
x
va
3
2
<
≤
x
ni olamiz;
b)
Yechish. x
+ 1 ning qiymatlari va bundan
x
ning qiymatlari ham butun
bo’lishi zarur. Bu qiymatlarda 3
x
2
–
x
ham butun bo’ladi va berilgan tenglama 3
x
2
–
x
=
x
+ 1 teng kuchli bo’ladi, bundan
x
= 1 ni olamiz.
c)
Yechish.
Berligan tenglamani 0
≤
x
< 4 qiymatlar qanoatlantiradi, bu qiy-
matlarda
x
4
3
butun qiymatlarni qabul qiladi, ya’ni
;
3
2
2
;
3
1
1
;
0
=
x
.
1
;
0
)
=
x
d
70.
.
11450
786
10
786
10
6
7
=
−
71.
Yechish.
.
686
7
5
999
7
999
5
999
999
=
+
−
−
72.
Yechish.
.
33
6
100
3
100
2
100
100
=
+
−
−
66
73.
98.
74
.
488.
75.
B
(2311; 5, 7, 13, 17) = 1378;
76.
B
(110; 2,3) = 37.
77
.
B
(12317; 3,5,7) = 5634.
78
. 393.
79
.
1
1
1
...
...
1
1
2
−
−
=
+
+
+
=
+
+
+
+
−
−
p
p
p
p
p
p
p
p
p
p
p
p
n
n
n
n
n
n
n
n
80
.
.
29
23
19
17
13
11
7
5
3
2
!
30
)
;
23
19
17
13
11
7
5
3
2
!
25
)
;
19
17
13
11
7
5
3
2
!
20
)
;
13
11
7
5
3
2
!
15
)
;
7
5
3
2
!
10
)
2
2
4
7
14
26
2
3
6
10
22
2
4
8
18
2
3
6
11
2
4
8
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
=
e
d
c
b
a
81.
.
19
17
13
11
2
!
10
!
10
!
20
2
⋅
⋅
⋅
⋅
=
82
.
Yechish.
,
7
!
100
!
1000
α
=
N
bu yerda
α
α
+
+
=
+
+
49
100
7
100
301
1000
49
1000
7
1000
shartni qanoatlantiradi, bundan
α
= 148 kelib chiqadi.
83
.
Yechish.
(
) (
)
( )
(
)
.
2
!
!
1
2
!
!
2
!
1
2
!
!
1
2
m
m
m
m
m
m
+
=
+
=
+
Agar
p
> 2 bo’lsa, u holda
∑
∑
=
=
+
<
+
≤
−
+
k
i
k
i
k
k
i
i
p
m
p
ерда
бу
p
m
p
m
1
1
1
1
2
,
1
2
.
84
.
Yechish.
Ixtiyoriy butun
x
=
k
(
a
≤
k
≤
b
) abssissa uchun [
f
(
x
)]+1 butun or-
dinatali va berilgan trapesiyaning ichida va chegarasida joylashadi. Demak, nuqtalar
soni
( )
(
)
∑
+
=
b
a
k
k
f
1
]
[
ga teng.
85.
126.
86.
Yechish.
Shartga asosan,
a
= 4
q
+ 1 yoki
a
= 4
q
+ 3 ga teng. Birinchi holda
(
)
.
2
1
3
6
3
2
4
3
4
2
4
−
=
=
+
+
=
+
+
a
q
q
q
q
a
a
a
Ikkinchi hol ham xuddi shunday
tekshiriladi.
87.
Yechish. a
=
mq
+
r
bo’lsin, bu yerda 0
≤
r
<
m
va (
r
,
m
) = 1. (
r, m
) = 1
shart barcha
m
≤
2 lar uchun bajarilishidan
r
= 1 kelib chiqadi. Demak,
(
) (
)(
)
∑
∑
−
−
=
−
⋅
−
=
=
+
−
=
−
=
1
1
1
1
.
2
1
1
2
1
1
1
m
i
m
i
m
a
m
m
m
a
i
q
m
q
i
67
88.
Yechish.
[ ]
1
0
,
<
≤
+
=
α
α
x
x
bo’lganligidan
[ ]
[ ]
,
2
1
2
2
1
+
+
=
+
+
α
x
x
x
bundan
2
1
1
2
1
2
1
<
+
≤
α
va
+
2
1
α
0 ga yoki 1 ga
teng.
[ ]
α
2
2
2
+
=
x
x
va
]
2
[
]
[
2
]
2
[
α
+
=
x
x
bo’lganligidan
+
+
−
=
+
+
2
1
]
2
[
]
2
[
2
1
]
[
α
α
x
x
x
.
Bu yerda
[ ]
,
0
2
2
1
=
=
+
α
α
yoki
[ ]
.
1
2
2
1
=
=
+
α
α
89
.
Yechish.
Tenglamaning har bir qismini
y
deb belgilab,
(
) (
)
.
1
1
бундан
,
1
1
+
−
<
≤
+
<
−
<
≤
y
m
x
my
оламиз
ни
y
m
x
m
x
y
Bu tengsizlikni qa-
noatlantiruvchi
x
lar mavjud bo’lishi uchun
my
< (
m
-1) (
y
+ 1) yoki
y
<
m
– 1 shartlar bajarilishi zarur va yetarli. Bundan quy-
ilagi natija kelib chiqadi:
my
≤
x
<(
m
- 1) (
y
+ 1), bu yerda
y
–
y
<
m
– 1 shartni qanoatlantiruvchi butun son.
90
.
Yechish.
ax
2
+
bx
+
c
funksiya va shu bilan birgalikda
[
ax
2
+
bx
+
c
] funksiya
a
> 0 da quyidan,
a
< 0 da esa yuqoridan chegaralangan.
Ikala holda ham [
ax
2
+
bx
+
c
] funksiya chegarasi
.
4
4
2
−
−
a
ac
b
songa teng. Shu sa-
babli
a
> 0 da berilgan tenglama yechimga ega bo’ladi, agar
;
4
4
2
d
a
ac
b
≤
−
−
shart
bajarilsa va faqat shartda, agar
a
< 0 bo’lsa, bu shart quyidagicha:
.
1
4
4
2
≥
−
−
a
ac
b
91
.
.
2
1
)
;
0
)
;
3
2
)
;
6
,
0
)
d
c
b
a
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