Matematika (informatika bilan) 1
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JIZZAX-2018
@alphraganus – matematik kanal MATEMATIKA (INFORMATIKA BILAN) 1. Agar , , uch xonali natural sonlar yig’indisi 777 ga teng bo’lsa, + + ni toping.
) )6 )8 )2 Yechimi: = 100 + 10 + = 100 + 10 + = 100 + 10 + + +
111 + + ): 777 = 1 + + ): 7 = 1 + + = 7
4,8 = + tenglikda va sonlar 5 dan kichik natural sonlar bo’lsa, y ning qiymatini toping. !)1 )3 #)$ )0
4,8 = + 4 + 0,8 = + 4 +
% = +
= 4 = 4
Javob: C)4 3. 2 < < 6 va 2 < < 10 bo’lsa, a va b butun sonlar uchun '( ) * '( * ) kasrning eng katta qiymatini toping. !) 7 3 +) , - )7 )15
1 + 1 +
= + + = 2 < < 6 va 2 < < 10 /01 2 33 4 25) = /01 2 33 4 25) /01 26 ℎ62 min)
= 5 3
Javob: +)
, -
4. Hisoblang: ;1 ' < = ∙ ;1
' ? = ∙ ;1 ' @ = ∙ … ∙ ;1 ' BC = )D ) 11 7 )7 ) 10 7
Yechimi: 8 7 ∙ 9 8 ∙
10 9 ∙ … ∙
63 62 =
63 7 = 9
Javob: )D 5. Besh xonali 734 sonini 55 ga bo’lganda natural son hosil bo’ladi. x ning barcha qiymatlari yig’indisini toping. )FF )9 )3 )14
734 soni 55 ga qoldiqsiz bo’linadi. U holda, bu son 5 va 11 ga ham qoldiqsiz bo’linadi. 5 ga bo’linish alomatiga ko’ra, oxirgi raqam nol yoki besh bo’lishi kerak. 11 ga bo’linish alomatiga ko’ra, toq o’rinda turgan raqamlar yig’indisi bilan juft o’rinda turgan raqamlar yig’indisining farqi nol yoki o’n birga karrali bo’lish kerak. 1) = 0 7340 + 3 + 0 − 7 + 4) = 0 H = I 2) = 5
7345 + 3 + 5 − 7 + 4) = 0 H = - 3 + 8 = 11 Javob:A)11 6.Hisoblang: ' C + C J + J C + ⋯ + ' C + 'B J !)72 )24 )65 L),M Yechim: N 1 2 + 3 2 + 5 2 + ⋯ +
15 2 O + N
2 3 +
4 3 + ⋯ +
16 3 O = 56
Javob:D)56 7. C + ) + C + 2 ) + ⋯ + C + 19 ) = 1425 tenglamani qanoatlantiruvchi x natural sonni toping.
!)6 )10 #), )8 Yechim: Ifoda arifmetik progressiyani tashkil etadi. ' =
+ 0 = 19
'@ = C + 19 C + + C + 19
2 ∙ 19 = 1425 C + 10 = 75 C + 10 − 75 = 0 = 5
31P ∙ 31Q + 31P + 31Q) ∙ 31 P + Q) !) − 1 +)F )2 )0
31 P + Q) = 1 − 31P ∙ 31Q 31P + 31Q
JIZZAX-2018
@alphraganus – matematik kanal 31P ∙ 31Q + 31P + 31Q) ∙ 1 − 31P ∙ 31Q 31P + 31Q =
= 31P ∙ 31Q + 1 − 31P ∙ 31Q = 1 Javob:B)1
5601 °
° + 5603
° + ⋯ + 560359 °
° L)S
Yechim: 560 + 560 = 0 1 T + = 360 U
° + 560359
° = 5602
° + 560358
° = 5603 ° + 560357 = ⋯ = 560180 ° = 0
Javob:D)0 10.Agar < −2 bo’lsa, V C + 6 + 1 + √9 − 12 + 4 C ifodani soddalashtiring. !)2 − ) + 2 #) − H − X ) − 2 Yechim: V9 − 12 + 4 C = V 3 − 2 ) C = |3 − 2 | < −2 |3 − 2 | = 3 − 2 V C
C + 4 + 4 = = V + 2) C = | + 2| < −2 | + 2| = − + 2) = − − 2 Javob:C) −H − X 11. Agar 2 Z = 81, 3 [ = 8 bo’lsa, ∙ ning qiymatini toping. !)14 +)FX )11 )13 Yechim: 2 Z = 81, \]1 C 81 = 3 [ = 8, \]1 J 8 =
∙ = \]1 C 81 ∙ \]1 J 8 = \]1
C 8 ∙ \]1
J 81 = 12
Javob:B)12 12.Ifodani soddalashtiring: Z ^ _'UZ ` ('B@ Z ` (BZ('J !) C − 5 + 13 ) C + 13 #)a X − Ma + F- ) C − 3 + 13 Yechim: % − 10 C + 169 =
% + 26
C − 36
C + 169 =
C + 13)
C − 36
C = C − 6 + 13) C − 6 + 13) 2 − 6 + 13
) 2 − 6 + 13 ) C + 6 + 13 = C − 6 + 13 Javob:C) a X − Ma + F-
@b `
@ = + ) C
tenglik ayniyat bo’ladi? ) −
F - ) − 1 ) − 1 4 ) − 1 2
c 3 − 1)
C 3 C d = + )
C
N 3 − 1 3 O
C = + ) C
N − 1 3O C = + )
C
= − 1 3
Javob: ) −
F -
C − 2 + 1) = C + 2 − 3 tenglama a ning qanday qiymatida cheksiz ko’p yechimga ega?
!) = −3 ) = 1, = −3 #)a = F ) ≠ 1 Yechim: f C − 2 + 1 = 0 C + 2 − 3 = 0 g h f ' = 1 C = −3,
J = 1
g h = 1 Javob: #)a = F 15.k ning qanday eng kichik natural qiymatida C + 2 + 2) C + 22 − 4 = 0 tenglamaning ildizlari 2 dan kichik bo’ladi? !)4 )3 )2 L)F
f ' + C = − 2 + 2) C ' C
= 22 − 4 g '
' − 2 < 0 C < 2, C − 2 < 0 ' − 2)
C − 2) > 0 ' C − 2
' + C ) + 4 > 0 22 − 4 + 2 2 + 2) C + 4 > 0
2 C + 52 + 4 > 0 2 + 4) 2 + 1) > 0 2 < −4; 2 > −1 Javob: L)F 16. 3 − ) + 2) > 0 tengsizlikning butun yechimlari yig’indisini toping. !) − 3 +)X )0 ) − 5 Yechim: 3 − ) + 2) > 0 −2 < < 3 −1 + 0 + 1 + 2 = 2 Javob: +)X 17. Agar
k ) = l − + 2, < 2 b_' C
g bo’lsa, k k −1)) ni toping. !) − 1 )3 #)F ) − 2
= −1, < 2 k −1) = − −1) + 2 = 3 = 3, ≥ 2 JIZZAX-2018
@alphraganus – matematik kanal k 3) =
3 − 1 2 = 1
Javob: #)F 18. Agar k ) =
+ − 4) ∙ J + 2 C + − 1) ∙ juft funksiya berilgan bo’lsa,
k ) ning qiymatini
toping. !)12 )14 )20 L)FI Yechim: n + − 4 = 0 − 1 = 0 g
C
k 3) = 2 ∙ 3 C = 18
Javob: L)FI 19. Hisoblang: o ;/
b + ' b = p .
C '
!)/ C + / − \02 +)r X − r + stX )/ C
C − / − \02 Yechim: u N/
b + 1 O p = g/ b + \0 | C ' 2 1 = / C + \02 − / ' − \01
= / C − / + \02 Javob: +)r
X − r + stX 20. o Jvb b∙wxCb ni hisoblang. !)3\02 + )6\0\02 + )1,5\0\02 + L)-ststXH + # Yechim: u 3p ∙ \02 = y \02 = 3
1 p = p3z = u 3p3 3 = 3\03 + = 3\0\02 + Javob: L)-ststXH + # 21.ABC uchburchakning BC tomonida D nuqta olingan. Agar = 16, = 4 va ! = ! = 10 bo’lsa, ADC uchburchak yuzini toping. )FX )14 )10 )16 Yechim: = { |}~ { |~• { |}~ = 16 ∙ €10
C − ;16
2 = C 2 = 48 16 4 = 48 { |~• { |~• = 12 Javob: )FX 22.To’g’ri burchakli ABCD trapetsiyaning B va C burchaklari to’g’ri, ! = 8,
= 6 va = 4. Trapetsiyaning D uchidan AC diagonaligacha bo’lgan masofani toping. !)3,6 )3 #)X, $ )2
! = V!
C + C = √64 + 36 = 10 { |}•~ = ! +
2 ∙ = 8 + 4 2 ∙ 6 = 36
{
= ! ∙
2 = 8 ∙ 6 2 = 24
{ |}•~ = {
|}• + {
|•~
36 = 24 + { |•~
{ |•~ = ! ∙ 2
12 = 10 ∙ 2
= 2,4 x - D uchidan AC diagonaligacha bo’lgan masofa.
Javob: #)X, $
asoslari DC=6, AB=2. BC tomonidan E nuqta olingan bo’lib,
BE=2EC bo’lsa,
ADE uchburchak yuzini toping. !)32 )18 )24 L)XI
Trapetsiyaning yon tomoni 2:1 nisbatda bo’linishi, uning balandligi ham xuddi shu nisbatda bo’linishini ifodalaydi. Agar ABCD trapetsiyaning balandligi h bo’lsa, u holda ABE va DEC uchburchaklarning balandliklari mos ravishda C• J
• J ga teng bo’ladi. { |}•~
= ! +
) ∙ ℎ 2 = 2 + 6) ∙ ℎ 2 = 4ℎ = 48 ℎ = 12 { |}‚ = ! ∙ 2ℎ
3 2 = 2 ∙ 2ℎ 6 =
2 ∙ 12 3 = 8
{ ~‚• = ∙ ℎ3
2 = 6 ∙ 12
6 = 12
{ |}•~ = {
|}‚ + {
~‚• + {
|~‚
48 = 8 + 12 + { |~‚
{ |~‚ = 28
Javob: L)XI 24. ABC uchburchak uchlarining koordinatalari berilgan: ! 8; 12), −8; 0) va −2; 8).
Uchburchakning CM medianasi yotgan to’g’ri chiziq tenglamasini tuzing. !) + 2 + 3 = 0 ) + + 6 = 0 #)H + ƒ = M ) − − 6 = 0 Yechim: M nuqta
AB tomonning o’rtasida. „ ;
b … (b † C ; … ( † C = = „ ;
?( _?) C ; 'C(U C = = „ 0; 6) − ‡ ‡ − • = −
‡
‡ − •
− 0 0 − −2) = − 6 6 − 8
JIZZAX-2018
@alphraganus – matematik kanal + = 6
Javob: #)H + ƒ = M 25. ! = { : | − 2| < 3, ‰Š} to’plamning elementlari sonini toping. !)3 +)$ )6 )5 Yechim: | − 2| < 3 −3 < − 2 < 3 −1 < < 5 = 1; 2; 3; 4
+)$ Testlar yechilishi davomida yo’l qo’yilgan xatolar uchun uzr!
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