= 1 tenglamani yeching.
tenglamani yeching.
tenglamani yeching.
yeching.
@Matematika
Harbiy
TEST 5. 07. 2019
https://t.me/Matematika
7
Qo’shko’pir 2019
M. Xudayberganov
https://t.me/Riyoziyot
153.
8
2𝑙𝑔𝑥
∙ 4
𝑙𝑔𝑥
= 128 tenglamani yeching.
A)
10
7
8
B)
10
7
C)
10
−
7
8
D)
10
1
8
154.
𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 = 2𝑡𝑔𝑥 + 1 tenglamani
yeching.
A)
𝜋𝑛, 𝑛 ∈ 𝑍
B)
𝜋𝑛,
𝜋
4
+ 𝜋𝑛, 𝑛 ∈ 𝑍
C)
−
π
4
+ 𝜋𝑛, 𝑛 ∈ 𝑍
D)
𝜋𝑛, −
π
4
+ 𝜋𝑛, 𝑛 ∈ 𝑍
155.
cos(12𝑎𝑟𝑐𝑡𝑔𝑥) = 1 tenglama nechta ildizga
ega?
A)
6 B) 1 C) 3 D) 5
156.
sin(12𝑎𝑟𝑐𝑡𝑔𝑥) = 0 tenglama nechta ildizga
ega?
A)
10 B) 11 C) 12 D) 5
157.
𝐸𝐾𝑈𝐵(𝑥; 4) = 1 bo’lsa, [
𝑥
4
] + [
2𝑥
4
] + [
3𝑥
4
] = 9
tenglama nechta ildizga ega? Bunda [
𝑎] − 𝑎
sonining butun qismi.
A)
0 B) 1 C) 2 D) 3
158.
𝐸𝐾𝑈𝐵(𝑥; 3) = 1 bo’lsa, [
𝑥
3
] + [
2𝑥
3
] = 9
tenglama nechta ildizga ega? Bunda [
𝑎] − 𝑎
sonining butun qismi.
A)
0 B) 1 C) 2 D) 3
159.
−1 + 2 − 3 + 4 − 5 + ⋯ + 198 − 199 ni
hisoblang.
A)
−99 B) −100 C) −101 D) 100
160.
−1 + 2 − 3 + 4 − 5 + ⋯ + 288 − 289 ni
hisoblang.
A)
−146 B) −144 C) −145 D) 145
161.
Agar arifmetik progressiyada dastlabki
20 ta
hadi yig’indisi
400 ga, dastlabki 30 ta hadi
yig’indisi esa
900 ga teng. Shu progressiyaning
dastlabki
50 ta hadi yig’indisini toping.
A)
1600 B) 2400 C) 2500 D) 2560
162.
Arifmetik progressiyada
𝑎
1
= 1, 𝑆
20
− 𝑆
12
=
380 bo’lsa, 𝑑 ni toping.
A)
5 B) 4 C) 2 D) 3
163.
Arifmetik progressiyada dastlabki
𝑛 ta hadi
yig’indisi
𝑆
𝑛
= 𝑛
2
+ 9𝑛 formula bilan
aniqlanadi. Shu progressiyaning
20 − hadini
toping.
A)
48 B) 30
C)
39 D) To’g’ri javob berilmagan
164.
Arifmetik progressiyada dastlabki
𝑛 ta hadi
yig’indisi
𝑆
𝑛
= 𝑛
3
+ 2𝑛
2
formula bilan
aniqlanadi. Shu progressiyaning
10 − hadini
toping.
A)
250 B) 125
C)
309 D) To’g’ri javob berilmagan
165.
Masshtabi
1 ∶ 4000000 bo’lgan xaritada ikki
shahar orasidagi masofa
1,2 dm ga teng bo’lsa,
bu ikki shahar orasidagi masofa masshtabi
1 ∶ 3000000 bo’lgan xaritada necha sm ga teng
bo’ladi?
A)
1,6 B) 1,8 C) 18 D) 16
166.
Masshtabi
1 ∶ 6000000 bo’lgan xaritada ikki
shahar orasidagi masofa
1,8 dm ga teng bo’lsa,
bu ikki shahar orasidagi masofa masshtabi
1: 4000000 bo’lgan xaritada necha sm ga teng
bo’ladi?
A)
2,7 B) 2,4 C) 24 D) 27
167.
Masshtabi
1 ∶ 4000000 bo’lgan xaritada ikki
shahar orasidagi masofa
1,2 dm bo’lgan
kesmaning haqiqiy uzunligi necha km?
A)
48 B) 240 C) 480 D) 24
168.
Masshtabi
1 ∶ 3000000 bo’lgan xaritada ikki
shahar orasidagi masofa
24 sm uzunlikdagi
kesma
1 ∶ 5000000 masshtabli xaritada qanday
dm uzunlikka teng bo’ladi?
A)
10,8 B) 14,4 C) 16 D) 12
169.
Agar qo’shni burchaklar
13 ∶ 17 nisbatda
bo’lsa, ulardan kichigini toping.
A)
52° B) 85° C) 65° D) 78°
170.
Agar qo’shni burchaklar
11 ∶ 19 nisbatda
bo’lsa, ulardan kichigini toping.
A)
56° B) 66° C) 36° D) 76°
171.
Agar qo’shni burchaklar
11 ∶ 19 nisbatda
bo’lsa, ulardan kattasini toping.
A)
124° B) 114° C) 144° D) 104°
172.
C nuqta AB kesmani A uchidan boshlab
hisoblaganda
4 ∶ 3 kabi, D nuqta esa AC
kesmani A uchidan boshlab hisoblaganda
5 ∶ 3
kabi nisbatda bo’ladi. Agar AB kesma
uzunligi
56 bo’lsa, DC kesma uzunligini toping.
A)
9 B) 15 C) 12 D) 8
173.
C nuqta AB kesmani A uchidan boshlab
hisoblaganda
2 ∶ 3 kabi, D nuqta esa AC
kesmani A uchidan boshlab hisoblaganda
3 ∶ 5
kabi nisbatda bo’ladi. Agar AB kesma uzunligi
44 bo’lsa, DC kesma uzunligini toping.
A)
6,6 B) 17,6 C) 11 D) 26,4
174.
𝐴𝐵 kesmada 𝐶 nuqta shunday tanlanganki,
𝐴𝐶 ∶ 𝐵𝐶 = 3 ∶ 4. 𝐴𝐶 kesmada 𝐷 nuqta shunday
tanlanganki,
𝐴𝐷 ∶ 𝐷𝐶 = 5 ∶ 3 shartlar o’rinli.
Agar
𝐴𝐵 kesma uzunligi 56 bo’lsa, 𝐷𝐶 kesma
uzunligini toping.
A)
9 B) 15 C) 24 D) 8
175.
Uchburchakning ikki tomoni mos ravishda 8 va
5 ga teng. Agar uchburchakning uchinchi
tomoni butun son bo’lsa, uchburchak
@Matematika
Harbiy
TEST 5. 07. 2019
https://t.me/Matematika
8
Qo’shko’pir 2019
M. Xudayberganov
https://t.me/Riyoziyot
perimetrining eng kichik qiymatini toping.
A)
17 B) 18 C) 25 D) 16
176.
To’g’ri burchakli uchburchakning
gipotenuzasiga tushirilgan balandligi ...
A) katetlaridan katta
B) katetlarining gipotenuzadagi proyeksiyalari
o’rta proporsionalining
0,25 qismiga teng
C) gipotenuzaning yarmiga teng
D) uchburchakni o’ziga o’xshash ikkita
uchburchakka ajratadi.
177.
To’g’ri burchakli uchburchakning
gipotenuzasiga tushirilgan balandligi ...
A) katetlaridan kichik
B) katetlarining gipotenuzadagi proyeksiyalari
o’rta proporsionalining
0,25 qismiga teng
C) gipotenuzaning yarmiga teng
D) uchburchakni ikkita o’xshash va tengdosh
uchburchaklarga ajratadi.
178.
To’g’ri burchakli uchburchakning
gipotenuzasiga tushirilgan balandligi ...
A) katetlaridan katta
B) katetlarining gipotenuzadagi proyeksiyalari
o’rta proporsionaliga teng
C) gipotenuzaning yarmiga teng
D) uchburchakni ikkita o’xshash va tengdosh
uchburchaklarga ajratadi.
179.
𝐴𝐵𝐶 to’g’ri burchakli uchburchakda 𝐶 to’g’ri
burchak. AN bissektrisa o’tkazilgan. Agar
𝐶𝑁 = 4, 𝐴𝐵 + 𝐴𝐶 = 14 bo’lsa, 𝐴𝐵𝐶
uchburchak yuzini toping.
A)
14 B) 18 C) 28 D) 56
180.
𝐴𝐵𝐶 to’g’ri burchakli uchburchakda 𝐶 to’g’ri
burchak. AN bissektrisa o’tkazilgan. Agar
𝐶𝑁 = 2𝑝, 𝐴𝐵 + 𝐴𝐶 = 𝑚 bo’lsa, 𝐴𝐵𝐶
uchburchak yuzini toping.
A)
2𝑚𝑝 B)
𝑚
2
−𝑝
2
4
C)
1
2
𝑚𝑝 D) 𝑚𝑝
181.
ABC uchburchakda |
𝐵𝐶| = 𝑎, |𝐴𝐶| = 𝑏,
| 𝐴𝐵| = 𝑐, 3 ∙ ∠𝐴 + ∠𝐵 = 180°, 3𝑎 = 2𝑐
bo’lsa,
𝐴𝐶 tomon uzunligini 𝑎 orqali ifodalang.
A)
𝑎√2 B) 𝑎√3 C)
5𝑎
4
D)
3𝑎
4
182.
ABCD teng yonli trapetsiyaning BC kichik
asosi
6 sm ga teng. B nuqtadan CD va AD
tomonlariga mos ravishda BK va BH
perpendikulyarlar tushirilgan. Agar BK=
3 sm
bo’lsa,
∠𝐵𝐴𝐻 ni toping.
A)
30° B) 60° C) 45° D) 75°
183.
ABCD teng yonli trapetsiyaning BC kichik
asosi
6 sm ga teng. B nuqtadan CD va AD
tomonlariga mos ravishda BK va BH
perpendikulyarlar tushirilgan. Agar BK=
3√3
sm bo’lsa,
∠𝐵𝐴𝐻 ni toping.
A)
30° B) 60° C) 45° D) 75°
184.
Trapetsiyaning bir diagonali
13 ga, ikkinchi
diagonali
√125 ga, balandligi esa 5 ga teng
bo’lsa, uning yuzini toping.
A)
55 B) 110 C) 25√5 D) 45
185.
Trapetsiyaning diagonallari
15 va 20 ga,
balandligi esa
12 ga teng bo’lsa, uning yuzini
toping.
A)
300 B) 150 C) 144 D) 225
186.
ABCD trapetsiya asoslari
𝐵𝐶 = 8 va 𝐴𝐷 = 32
ga teng. Agar
∠𝐴𝐷𝐶 = ∠𝐵𝐴𝐶 bo’lsa, AC
diagonalni toping.
A)
17 B) 18 C) 16 D) 20
187.
ABCD trapetsiya asoslari
𝐵𝐶 = 27 va 𝐴𝐷 =
48 ga teng. Agar ∠𝐴𝐷𝐶 = ∠𝐵𝐴𝐶 bo’lsa, AC
diagonalni toping.
A)
45 B) 36 C) 18 D) 27
188.
ABCD trapetsiya asoslari
𝐵𝐶 = 𝑏 va 𝐴𝐷 = 𝑎
ga teng. Agar
∠𝐴𝐷𝐶 = ∠𝐵𝐴𝐶 bo’lsa, AC
diagonalni toping.
A)
1
2
(𝑎 + 𝑏) B) √𝑎
2
+ 𝑏
2
C)
2√𝑎𝑏 D) √𝑎𝑏
189.
ABCD to’g’ri burchakli trapetsiyada A va D
burchaklar to’g’ri. Tomonlari
𝐷𝐶 = 4, 𝐴𝐵 =
𝐵𝐶 = 10 bo’lsa, shu trapetsiya yuzini toping.
A)
48 B) 56 C) 28 D) 63
190.
ABCD to’g’ri burchakli trapetsiyada A va D
burchaklar to’g’ri. Tomonlari
𝐷𝐶 = 8, 𝐴𝐵 =
𝐵𝐶 = 13 bo’lsa, shu trapetsiya yuzini toping.
A)
64 B) 92 C) 126 D) 136
191.
𝑎⃗(−1; 2), 𝑏⃗⃗(−2; 1), 𝑐⃗(−3; 2) hamda 2𝑎⃗ − 𝑘𝑏⃗⃗
va
𝑐⃗ vektorlar perpendikulyar bo’lsa, 𝑘 ni
qiymatini toping.
A)
4
7
B)
7
4
C)
−
7
4
D)
−
1
2
192.
𝑎⃗(−1; 2), 𝑏⃗⃗(−2; 1), 𝑐⃗(−3; 2) hamda 𝑎⃗ − 2𝑘𝑏⃗⃗
va
−2𝑐⃗ vektorlar perpendikulyar bo’lsa, 𝑘 ni
qiymatini toping.
A)
−
16
7
B)
7
16
C)
−
7
16
D)
−
1
8
193.
O’nbirburchakli prizmaning nechta turli
diagonal kesimi mavjud?
A)
44 B) 54 C) 35 D) 33
194.
Beshburchakli prizmaning nechta turli diagonal
kesimi mavjud?
A)
10 B) 3 C) 2 D) 5
195.
O’nbirburchakli piramidaning nechta turli
diagonal kesimi mavjud?
A)
44 B) 54 C) 35 D) 33
@Matematika
Harbiy
TEST 5. 07. 2019
https://t.me/Matematika
9
Qo’shko’pir 2019
M. Xudayberganov
https://t.me/Riyoziyot
196.
Oltiburchakli prizmaning nechta turli diagonal
kesimi mavjud?
A)
18 B) 9 C) 6 D) 15
197.
Ko’pyoqning yoqlari soni 7 ta, uchlari soni 6 ta
bo’lsa, undagi qirralar soni nechta?
A)
10 B) 11 C) 12 D) 14
198.
O’q kesimi kvadratdan iborat bo’lgan silindr
asosida uzunligi
3 ga teng bo’lgan vatar 60° li
yoyni tortib turadi. Silindr hajmini toping.
A)
27√3𝜋 B) 54𝜋 C) 48𝜋 D) 27√2𝜋
199.
O’q kesimi kvadratdan iborat bo’lgan silindr
asosida uzunligi
1 dm ga teng bo’lgan vatar 60°
li yoyni tortib turadi. Silindr hajmini (
𝑑𝑚
3
)
toping.
A)
√3𝜋 B) √2𝜋 C) 2𝜋 D) 2√2𝜋
200.
O’q kesimi kvadratdan iborat bo’lgan silindr
asosida uzunligi
2√3 ga teng bo’lgan vatar 60°
li yoyni tortib turadi. Silindr hajmini toping.
A)
48√3𝜋 B) 24√3𝜋 C) 72𝜋 D) 48√2𝜋
Informatika
201.
25
6
+ 27
8
− 103
4
= 𝑥
7
bo’lsa,
𝑥 ni toping.
A)
30 B) 21 C) 32 D) 25
202.
12
4
+ 21
4
+ 𝑥
4
= 45
8
bo’lsa,
𝑥 ni toping.
A)
10 B) 112 C) 22 D) 211
203.
15
6
+ 12
4
− 𝑥
5
= 16
10
bo’lsa,
𝑥 ni toping.
A)
11 B) 1 C) 113 D) 5
204.
103
4
+ 210
5
= 𝑥
10
bo’lsa,
𝑥 ni toping.
A)
74 B) 84 C) 54 D) 64
205.
Axborotni hajmi
32
𝑥+2
Kb yoki
256
𝑥
Mb
bo’lsa,
𝑥 ni qiymatini toping.
A)
8 B) 0 C)
10
3
D)
0, (3)
206.
{
𝐸𝐾𝑈𝐵(𝑥
9
; 𝑦
9
) = 45
9
𝑥
𝑦
=
11
10
7
10
bo’lsa,
𝑥𝑦 ko’paytmani
qiymatini 9 lik sanoq sistemasida hisoblang.
A)
216478 B) 261488
C)
216488 D) 214688
207.
𝐹𝐹𝐹𝐹
16
∙ 𝐴𝐴𝐴
16
= (? )
16
A)
𝐴𝐴9𝐹556 B) 𝐴𝐵8𝐸466
C)
𝐴𝐶7𝐷536 D) 𝐴𝐷6𝐶476
208.
Arifmetik progressiyada hadlari
𝑎
1
= 2
3
,
𝑎
2
=
10
3
va
𝑎
𝑛
= 1002
3
bo’lsa, uning dastlabki
𝑛
ta hadi yig’indisini toping.
A)
434 B) 424 C) 361 D) 868
209.
∫ 𝑑𝑥
𝐹
𝐵
ni 16 lik sanoq sistemasida hisoblang.
A)
4 B) −4 C) 8 D) 2
210.
{
104
5
+ 2𝑥
5
= 𝑦
6
+ 1
10
2
+ 3𝑦
5
= 𝑥
6
+ 1
sistemadan
𝑥 va 𝑦 ni
toping.
A) (
3; 4) B) (3; 3) C) (2; 3) D) ∅
211.
{
14
5
+ 2𝑥
5
= 𝑦3
6
+ 1
10
2
+ 3𝑦
5
= 𝑥1
6
+ 1
sistemadan
𝑥 va 𝑦 ni
toping.
A) (
3; 4) B) (3; 3) C) (2; 3) D) ∅
212.
𝐴(10; 14)
15
; 𝐵(7; 5)
8
; 𝐶(3; 9)
10
uchlarining
koordinatalari berilgan
uchburchakning yuzini
toping.
A)
42 B) 44 C) 45 D) 54
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