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JIZZAX-2018
@alphraganus – matematik kanal MATEMATIKA FANIDAN 2018 YIL IMTIHON SAVOLLARI 1. Ifodani soddalashtiring: ∈ (−3; 1) (1 − 2 + )( − 1)( − 1): √ + 1
+ 2 − 3
( − 1) ( + 1): √ + 1
( + 3)( − 1) = = ( − 1) ( + 1) ∙ ( + 3)( − 1) √ + 1
= = | − 1| ( + 1) ∙ ( + 3)( − 1) √ + 1
= = (1 − )( + 3)( − 1) = −( + 3)( − 1)
400 ∙
23 − 17 0,6
$
400 ∙ 23 − 17
0,6 $ = %400 ∙ √400 $ = √8000
$ = 20
3. Hisoblang: (2 &' + 1) − (2 &' − 1) Yechim: (2 &' + 1) − (2 &' − 1) = (2 &' + 1) − (2 &' − 1) = 2 4. Hisoblang: 12 25 ∙ 244 15 ∙ (38 − 23 ) )
12 25 ∙ 244 15 ∙ (38 − 23 ) ) =
25 ∙
244 15 ∙ 61 ∙ 15 ) = = 12 25 ∙ 4 15 ∙ 15
) =
12 25 ∙
2 15 ) = 2 5 5.Hisoblang: 2 ' + 2 && + 1 − 2 ' − 2
& + 4
Yechim: 2 ' + 2 && + 1 − 2 ' − 2
& + 4 = (2
&' + 1) −
− (2 &' − 2) = 2 &' + 1 − 2
&' + 2 = 3
6.Hisoblang: 4 * + 2 ∙ 6 * + 9 * − 4
, + 6
* + 9
*
4 *
* + 9
* − 4
, + 6
* + 9
*
= (2 * + 3
* ) −
− (2 , + 3 * ) = 2
* + 3
* − 2
, − 3
* = 128
7.Hisoblang: 4 &- + 6 ' + 9 ' − 4
&- − 6
' + 9
'
4 &-
' + 9
' − 4
&- − 6
' + 9
'
= (2 &- + 3
' ) −
− (2 &- − 3 ' ) = 2
&- + 3
' + 2
&- − 3
' = 2
'
8.Hisoblang: 4 & + 6 &. + 9 &. + 4
& − 6
&. + 9
&.
4 &
&. + 9
&. + 4
& − 6
&. + 9
&.
= (2 & + 3
&. ) −
− (2 & − 3 &. ) = 2
& + 3
&. + 2
& − 3
&. = 2
&.
9.Hisoblang: /√5 − √110(√33 + √15 − √22 − √10) √75 − √50
/√5 − √110 1√3 ∙ /√11 + √50 − √2 ∙ /√11 + √502 √75 − √50
=
5 ∙ /√3 − √20 = 5 − 11 5 = −1,2
10.Hisoblang: + % + √ + ⋯ $ $
= 2 − % − √ − ⋯ =?
√ + 2
$ = 2
+ 2 = 8 = 6
%2 − 2 − √2 − ⋯ = 5 (5 > 0) √2 − 5 = 5 2 − 5 = 5 5 = 1
11.Hisoblang: 7 8 9 − 7 + 7
: ; − 7 , < =? (7 < 0) Yechim: 7 − |7| + |7| − 7 = 0
5 + 1 + /√5 − 10 = 20 Yechim: 5 + 1 + 5 − 1 = 20 (5 ≥ 1) 25 = 20 5 = 10
JIZZAX-2018
@alphraganus – matematik kanal 13.Tenglamani yeching: 5 + 11 + 5 + 11 = 20 Yechim: 5 + 11 = ? ? + ? − 20 = 0 ? & = −5, ? = 4 5 + 11 = −5 @ ∅ 5 + 11 = 4 5 + 11 = 16 5 = ±√5
kattasidan 21 ga kam, o’rta geometrigi esa kichigidan 12 ga ko’p. Shu sonlarning kattasini toping. Yechim: C DEF = 5 − 21 5G = G + 12 H @ I G = 5 − 42 5G = G + 24G + 144 5=G+42G+42G=G2+24G+144 G + 42G = G + 24G + 144 18G = 144 G = 8
5 = 50
J KEL
+ K JEL + L JEK = 5 bo’lsa, M7 + N +
7 + N +
N + 7O : ( + 7 + N) =?
+ P Q R Q S J T KEL + JK JEL + JL JEK = 5
JK KEL
+ K T JEL + KL JEK = 75
JL KEL
+ KL JEL + L T JEK = N5
H 7 + N +
7 + N +
N + 7 + + 7 + N = 5( + 7 + N)
7 + N +
7 + N +
N + 7 = (5 − 1)( + 7 + N)
M7 + N +
7 + N +
N + 7O : ( + 7 + N) = 5 − 1
J KEL
+ K JEL + L JEK = 0 bo’lsa, M7 + N +
7 + N +
N + 7O : ( + 7 + N) =?
M7 + N +
7 + N +
N + 7O : ( + 7 + N) = 0 − 1 = −1
J KEL
+ K JEL + L JEK = 2 bo’lsa, M7 + N +
7 + N +
N + 7O : ( + 7 + N) =?
M7 + N +
7 + N +
N + 7O : ( + 7 + N) = 2 − 1 = 1 17. Agar J KEL + K JEL + L JEK = 3 bo’lsa, M7 + N +
7 + N +
N + 7O : ( + 7 + N) =?
M7 + N +
7 + N +
N + 7O : ( + 7 + N) = 3 − 1 = 2 18.Agar 7N = 5 bo’lsa, U 2
3 7 − NV U
4 N − 7V =?
U 2 − 7N V U 3 − 7N
7 V U 4 − 7N
N V =
(2 − 5)(3 − 5)(4 − 5) 5
= − 6 5 = −1,2
− 7 + 2 + 67 − 8 ko’phadning ko’paytuvchilardan birini toping. Yechim: + 2 + 1 − 7 + 67 − 9 = ( + 1) − (7 − 3) = ( + 1 − 7 + 3)( + 1 + 7 − 3) = ( − 7 + 4)( + 7 − 2)
+ I (5 − 5G + G )(5 + G ) = 25 (5 + 5G + G )(5 + G ) = 225 H (5 − 5G + G )(5 + G ) + (5 + 5G + G )(5 + G ) = 250 (5 − 5G + G + 5 + 5G + G )(5 + G ) = 250 2(5 + G )(5 + G ) = 250 (5 + G )(5 + G ) = 125 (5 + G )
. = 125
5 + G = 5 I(5 − 5G )5 = 25 5 + G = 5 H @ I 5G = 4 5 + G = 5 @ H (1; 2); (1; −2); (4; 1); (4; −1) 21.Tengsizlikni yeching: 5 |5 + 2| + 2 ≤ |5 + 2| − 2 Yechim: 5 |5 + 2| + 2 ≤ |5 + 2| − 2 |5 + 2| = 5 + 2 ≤ − 2 5 ≤
− 4 0 ≤
− 9 0 ≤ ( − 3)( + 3) 3 ≤ |5 + 2| ≥ 3 (5 + 2 − 3)(5 + 2 + 3) ≥ 0 (5 − 1)(5 + 5) ≥ 0 (−∞; −5Y ∪ [1; ∞)
27 − \]^ J . 7 . (1 + \]^ J 7 + \]^
K ) ∙ \]^
J 7 ∙ \]^ J _ 7 Yechim: \]^
J 7 = 5
JIZZAX-2018
@alphraganus – matematik kanal 27 − (35) . 11 + 5 + 152 ∙ (1 − 5) ∙ 1 95 = 27(1 − 5 . ) (1 + 5 + 5 ) 5 (1 − 5) ∙ 1 95 = 3 23.Hisoblang: ` abc 125 + d 32 e5
f '
Yechim: ` abc 125 + d 32 e5 = − 1 2 cos 125 + d 32 j
d 2 0 =H f ' − 1 2 cos 12 ∙ d 2 + d 32 +
1 2 cos 12 ∙ 0 + d 32 =
1 2
24.Hisoblang: ` 4 35 + 2 e5 & ' Yechim: ` 4 35 + 2 e5 & ' = 4 3 \c (35 + 2) k 1 0 =
4 3 \c10
H 25.Hisoblang: ` ?^ 5 + 2 abc5 e5
` M ?^ 5
abc5 + 2 abc5O e5 = ` abc5 N]a 5 e5 + 2 ` 1 abc5 e5
l & = ` abc5
N]a 5 e5 = ` 1 N]a 5 e(N]a5) = − 1 N]a5
l = 2 `
1 abc5 e5 = 2 \c m?^ 5 2m
l = l & + l = − 1 N]a5 + 2\c m?^ 5 2m + n
o(5) = −?^ 25, o p (5) =? Yechim: o p (5) = (−?^ 25) p = −2?^25 ∙ 2 1 N]a 25 =
= −
4?^25 N]a 25
o(5) = 5 8 − 55 − 1 funksiyaning [−1; 1Y oraliqdagi eng katta qiymatini toping. Yechim: o p (5) = 55 − 205 . = 0 55 . (5 − 4) = 0 5 = 0 5 ≠ 4
r(s) = −t o(1) = −5 o(−1) = −7
1 + ((N]au + N]av) + (abcu + abcv) ): N]a u − v
2
1 + ( N]a u + 2N]auN]av + N]a v + +abc u + 2abcuabcv + abc v): N]a u − v 2 =
1 + (2 + 2N]auN]av + 2abcuabcv): N]a u − v 2 =
1 + 2(1 + cos(u − v)): N]a u − v 2 =
1 + 4N]a
u − v 2 : N]a
u − v 2 = 5
\]^ D
13 = \]^ w.D
13 Yechim: 5 = 4 − 35 5 + 35 − 4 = 0 x t = − 5 ≠ 1
30.Soddalashtiring: 1 +
(N]au + N]av) (abcu − abcv)
1 +
U2N]a u + v 2 N]a
u − v 2 V
U2abc u + v 2 N]a
u − v 2 V
= 1 + N?^ u + v
2 =
= 1 abc u + v 2
31. Hisoblang: ` 3 25 − 1 e5 &
Yechim: ` 3 25 − 1 e5 & = 3 2 \c |25 − 1| k 2 1 =
3 2 \c3
H 32. Hisoblang: ` U5 +
1 5V e5
&
` (5 + 2 + 1 5 )e5 & = 5 . 3 + 25 −
1 5 k
2 1 =
29 6 H 33.Hisoblang: abc y
1 2 zNabc M− 2√2 3 O{
Yechim: −abc M
1 2 zNabc
2√2 3 O = −
1 − N]a M zNabc 2√2 3 O
2
− 1 − 13 2 = −
1 3 = −
1 √3
34. o(5) =
& L|}
T D , o p 1 f 2 =? Yechim: o p (5) = (−?^ 25) p = −2?^25 ∙ 2 1 N]a 25 =
= −
4?^25 N]a 25
JIZZAX-2018
@alphraganus – matematik kanal o p 1 d 22 = − 4?^d N]a d = 0
~ = {1; 2; 3; 5; 6; 7; 10} • = {6; 7; 8; 9; 10; 11} ~ ∩ • =?
~ ∩ • = {6; 7; 10}
abc
&'' 5 + N]a
&'' 5 = 1
Yechim: abc5 = ±1 @ N]a5 = 0 N]a5 = ±1 @ abc5 = 0 5 =
dc 2 , c ∈ ƒ
5 & = 2, 5 = 3, 5 .
25 . − 55 − 135 + c = 0 ushbu tenglamaning ildizlari bo’lsa, 5 . ni toping. Yechim: 5 & +5 + 5 . = 5 2
2 + 3 + 5 . = 5 2
5 . = − 5 2
38. I : + &' = 18 - + && = 38 H Š
&8 =? Yechim: − I
: + &' = 18 - + && = 38
H −3e − e = −20 e = 5 &
& + 9e = 18 2 &
& = −26
Š &8 = 2 & + 14e 2 ∙ 15 =
−52 + 70 2 ∙ 15 = 135 39. I & = 108 . = 180 H & =? Yechim: + I
& = 108
. = 180
H ( & + . ) = 288 2 = 288 = ±12
& = ±9
40.Tenglamani yeching: 2 }‹Œ T D + 2 L|} T D = 3 Yechim: abc5 = ±1 @ N]a5 = 0 N]a5 = ±1 @ abc5 = 0 5 =
dc 2 , c ∈ ƒ
41.Tenglamani yeching: |55 − 3| + |35 − 5| = 95 + 10 Yechim: •)5 < 0,6 −55 + 3 − 35 + 5 = 95 + 10 175 = −2 x = − Ž
mumkin. •)0,6 < 5 < 5 3
55 − 3 − 35 + 5 = 95 + 10 −75 = 8
5 = − * , mumkin emas. ‘)5 >
5 3
55 − 3 + 35 − 5 = 95 + 10 5 = −18 mumkin emas.
((N]au − N]av) + (abcu − abcv) ): U4abc u − v 2 V
Yechim: ( N]a u − 2N]auN]av + N]a v + +abc u − 2abcuabcv + abc v): U4abc u − v
2 V =
(2 − 2N]auN]av − 2abcuabcv): U4abc u − v 2 V =
2(1 − cos(u − v)): U4abc u − v 2 V
= 4abc
u − v 2 : U4abc u − v 2 V = 1
` (|5| + 1)e5 '
Yechim: ` (5 + 1)e5 ' =
2 + 5 m 2 0 = 2 + 2 − 0 = 4 H 44.Tenglamaning yechimlari soni nechta? 5 . = U 1 3V D + 1
Yechim: 1 ta.
JIZZAX-2018
@alphraganus – matematik kanal 45. ’ 5
- (5 8 + 1) Œ (5 8 − 1)
Πe5 & ' = ,
& J =? Yechim: ` 5
- ((5
8 + 1)(5
8 − 1))
Œ e5 & ' = ` 5 - (5 &' − 1) Œ e5 & ' = “ 5 &' = ?
105 - e5 = e?” = ` ?
Œ e? 10 = ? ŒE&
10(2c + 1) m 1 0 = H & ' 1 = 20c + 10 46. G = 45 va G = −45 − 8 parabolalarga abssissalar o’qi bilan o’tkir burchak tashkil qiladigan umumiy urinma o’tkazilgan. Shu urinmaning tenglamasini ko’rsating.
G = •5 + \, • > 0, I •5 + \ = 45 •5 + \ = −45 − 8 @ I 45 − •5 − \ = 0 45 + •5 + \ + 8 = 0 @ HH – = 0 @ − I • + 16\ = 0 • − 16(\ + 8) = 0 H @ —\ = −4 • = 8 H
47.a ning qanday eng kichik butun qiymatida −5 − 105 + 5 < tengsizlik x ning barcha qiymatlarida o’rinli bo’ladi?
5 + 105 + − 5 > 0 – < 0 100 − 4( − 5) < 0 25 − ( − 5) < 0 25 − + 5 < 0 30 < (30; ∞)
= 31
o(5) = 6 + 5?^ 25, o p (d) =? Yechim: o p (5) = (6 + 5?^ 25) p = 10?^25 ∙ 2 1 N]a 25 =
= 20?^25 N]a 25
o p (d) = 0
49. I : + &' = 38 - + && = 23 H Š
&8 =? Yechim: − I
: + &' = 38 - + && = 23
H −3e − e = 15 e = −3,75 & + 5e + & + 9e = 15 2 &
2 & = 67,5 Š &8 = 2 & + 14e 2 ∙ 15 =
67,5 − 52,5 2 ∙ 15 = 112,5 50.Tenglamani yeching: 5 + 3 − 4√5 − 1 = 1 Yechim: 5 − 1 + 4 − 4√5 − 1 = 1 √5 − 1 = − 4 + 3 = 0 = 1, = 3
√5 − 1 = 1 5 = 2 √5 − 1 = 3 5 = 10 Testlar yechilishi davomida yo’l qo’yilgan xato va kamchilik uchun uzr so’raymiz. Har qism 50 ta testlar ishlanmasidan iborat. Bu 1-qism. Download 229.48 Kb. Do'stlaringiz bilan baham: 
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