MATEMATIKA (INFORMATIKA BILAN)
1.
= 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ⋯ + 40 ∙ 41,
= 5 ∙ 4 + 10 ∙ 6 + 15 ∙ 8 + ⋯ + 200 ∙ 82
bo’lsa, ning qiymatini toping.
)
1
12 )
1
6 )
)
1
8
Yechim:
=
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ⋯ + 40 ∙ 41
5 ∙ 4 + 10 ∙ 6 + 15 ∙ 8 + ⋯ + 200 ∙ 82
=
1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ⋯ + 40 ∙ 41
5 ∙ 2 ∙ (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ⋯ + 40 ∙ 41)
=
1
10
Javob:
)
2.
!"
#
$%&
kasr ma’noga ega bo’lmaydigan
barcha
x
lar yig’indisini toping.
') ) − 1 ))1 ) − 2
Yechim:
7
1 + 2
+ − 1
=
7(+ − 1)
+ + 1 , + ≠ 1; + ≠ −1
−1 + 1 = 0
Javob:
')
3.Agar
+ < −1; 0 > 1 bo’lsa, quyidagi
javoblardan qaysi biri har doim o’rinli?
)+
2
> 0 3)4
5
> 6
5
))+
7
< 0
7
)0
7
> +
8
Yechim:
+-manfiy son. Manfiy sonning toq darajasi ham
manfiy bo’lib qolaveradi.
0-musbat son. Musbat sonning toq darajasi ham
musbat bo’lib qolaveradi.
Javob:
3)4
5
> 6
5
4. Hi
soblang:
91
!
: ∙ 91
!
;
: ∙ 91
!
<
: ∙ … ∙ 91
!
8<
:
)7 )
10
7 ))
69
7 ?)
Yechim:
8
7 ∙
9
8 ∙
10
9 ∙ ⋯ ∙
70
69 =
70
7 = 10
Javob:
?)
5.180 gramm suvga 70 gramm tuz aralashtirildi.
Hosil bo’lgan aralashmaning necha foizi tuzdan
iborat bo’ladi?
')@A )25 ))30 )22
Yechim:
70
180 + 70 ∙ 100% = 28%
Javob:
')@A
6.Agar
C + D + √ + ⋯
F
F
F
= 2
bo’lsa,
C − D − √ − ⋯ ning qiymatini toping.
)1 3)@ ))4 )3
Yechim:
G + C + √ + ⋯
F
F
F
= 2
√ + 2
F
= 2
+ 2 = 8
= 6
G − C − √ − ⋯ = G6 − C6 − √6 − ⋯ = +
√6 − + = +
6 − + = +
7
+
7
+ + − 6 = 0
+
!
= −3, 6
@
= @ + > 0
Javob:
3)@
7.Hisoblang:
1 ∙ 4 + 2 ∙ 7 + 3 ∙ 10 + ⋯ + 9 ∙ 28
')I )740 ))1210 )960
Yechim:
1 ∙ 4 + 2 ∙ 7 + 3 ∙ 10 + ⋯ + J ∙ (3J + 1) == J ∙ (J + 1)
7
1 ∙ 4 + 2 ∙ 7 + 3 ∙ 10 + ⋯ + 9 ∙ 28 =
9 ∙ (9 + 1)
7
= 9 ∙ 100 = 900
Javob:
')I
8.Agar
KLMN = −
!
7
bo’lsa,
LM3Nning qiymatini
toping.
) −
1
11 )5,5 ) −
@
)
1
6
Yechim:
KLMN = −
!
7
, LMN = −2
LM3N =
3LM
N
− LM
3
N
1 − 3LM
2
N
= 3 ∙
(
−2
)
− (−2)
3
1 − 3 ∙ (−2)
2
=
= − 2
11
Javob:
) −
@
9.Hisoblang:
OPJ2
°
+ OPJ3
°
+ OPJ4
°
+ ⋯ + OPJ358
°
)1 )OPJ179
°
) ) − 1
Yechim:
OPJ2
°
+ OPJ358
°
= 2OPJ
2
°
+ 358
°
2
KQO
2
°
− 358
°
2
= 0
OPJ+ + OPJ0 = 0 , + + 0 = 360
°
Javob:
)
10.Agar
+ < −2 bo’lsa,
D+
7
+ 5+ + 2 + √4 − 4+ + +
7
ifodani
soddalashtiring.
)2 + + )2 − + )) − 2+ ?) − 6 − @
Yechim:
C+
7
+ 5+ + 2 + D(+ − 2)
7
=
= D+
7
+ 5+ + 2 + |+ − 2| =
= D+
7
+ 5+ + 2 + 2 − + = D+
7
+ 4+ + 4
= D(+ + 2)
7
= |+ + 2|
= −+ − 2
Javob:
?) − 6 − @
11.
+, 0, S butun sonlar bo’lib, 0 < 0 va
7
TU
= −
T
2V
=
2
WX
bo’lsa,
+, 0, S sonlarini o’sish
tartibida joylashtiring.
)+ < 0 < S )S < 0 < +
)4 < + < S )0 < S < +
Yechim:
0 < 0, + > 0, S > 0
Sur’atni bir xil songa keltirib olamiz.
12
18+ = −
12
160 =
12
15S
18+ = −160 = 15S
6+ = 5S
Javob:
)4 < + < S
12.Ifodani soddalashtiring:
1 + 1
+ K
1 − 1
+ K
∙ Y1 +
7
+ K
7
−
7
2 K
Z :
( + + K)
7
K
)1 3) , \ )) + K − ) + + K
Yechim:
+ K +
∙ ( + K)
+ K −
∙ ( + K)
∙ Y
2 K +
7
+ K
7
−
7
2 K
Z ∙
K
( + + K)
7
=
+ K +
+ K − ∙ Y
( + K)
7
−
7
2 K
Z ∙
K
( + + K)
7
=
=
+ K +
+ K − ∙ Y
( + K + )( + K − )
2 K
Z ∙
K
( + + K)
7
= 0,5
Javob:
3) , \
13.Agar
+ ≠ 0 bo’lsa,
5 + 5
7U"V
− 5
U"!
− 5
U"V
= 0
tenglamadagi x ni y orqali ifodalang.
)+ = −1 − 0 3)6 = − 4
))+ = 0 − 1 )+ = 0 + 1
Yechim:
5 − 5
U"!
+ 5
7U"V
− 5
U"V
= 0
5 ∙ (1 − 5
U
) + 5
U"V
∙ (5
U
− 1) = 0
(1 − 5
U
) ∙ (5
U"V
− 5) = 0
5
U"V
− 5 = 0
5
U"V
= 5, + + 0 = 1
Javob:
3)6 = − 4
14.Agar
+√+ − 7√+ = 6 bo’lsa, + − √+ ning
qiymatini toping.
)7 3)] ))8 )3
Yechim:
+√+ − 7√+ − 6 = 0
+√+ − √+ − 6√+ − 6 = 0
√+ ∙ (+ − 1) − 6 ∙ ^√+ + 1_ = 0
√+ ∙ ^√+ − 1_^√+ + 1_ − 6 ∙ ^√+ + 1_ = 0
^√+ + 1_ ∙ (√+ ∙ ^√+ − 1_ − 6) = 0
^√+ + 1_ ∙ (+ − √+ − 6) = 0
+ − √+ = 6
Javob:
3)]
15.Agar
+
7
+ (` + 2)
7
∙ + + 2` − 4 = 0
tenglamaning ildizlari 2 dan kichik bo’lsa,
k
ning eng katta butun manfiy qiymatini toping.
) − 2 ) − 4 )) − 1 ?) − \
Yechim:
+
!
< 2, +
7
< 2
+
!
− 2 < 0, +
7
− 2 < 0
(+
!
− 2)(+
7
− 2) > 0
+
!
+
7
− 2(+
!
+ +
7
) + 4 > 0
a+
!
+ +
7
= −(` + 2)
7
+
!
+
7
= 2` − 4
b
2` − 4 + 2(` + 2)
7
+ 4 > 0
`
7
+ 5` + 4 > 0
(` + 4)(` + 1) > 0
` ∈ (−∞; −4) ∪ (−1; ∞)
` = −5
Javob:
?) − \
]. f
2g7U
!"TU
f > 0 tengsizlikni yeching.
) h−∞; −
1
3i ∪ (2;
∞
)
) h−∞; −
1
3i ∪ h−
1
3 ;
∞
i
) h−∞; − 5i ∪h−5;@i ∪(@;∞)
)(−
∞;
∞)
Yechim:
j
4 − 2+
1 + 3+j > 0
4 − 2+ ≠ 0 , + ≠ 2
1 + 3+ ≠ 0 , + ≠ −
1
3
Javob:
) h−∞; − 5i ∪h−5;@i ∪(@;∞)
17. Agar
k(2+ − 3) = 3+ + 5 bo’lsa, k(k(1))
ni toping.
)11 )38 )@] )16
Yechim:
2+ − 3 = 1, + = 2
k(2 ∙ 2 − 3) = 3 ∙ 2 + 5
k(1) = 11, 2+ − 3 = 11, + = 7
k^k(1)_ = k(11) = k(2 ∙ 7 − 3) = 3 ∙ 7 + 5 = 26
Javob:
)@]
18.
0 = KQO
7
9
U
T
−
l
2
: + 2OPJ+ funksiyaning eng
kichik musbat davrini toping.
)2m 3)]n ))3m )o pqP0 rs O
Yechim:
t
!
=
l
&
F
= 3m, t
7
= 2m
uvwv(t
!
; t
7
) = 6m
Javob:
3)]n
19.
+ = 1, 0 = r
U
va
0 = r
gU
funksiyalar bilan
chegaralangan soha yuzini toping.
')
(x − )
@
x
)r − 1 ))
r − 1
r )
(r − 2)
7
r
Yechim:
r
U
= r
gU
, + = 0
y (r
U
− r
gU
)o+ = r
U
+ br
gU
|
1
0 = r +
1
r − 1 − 1
!
z
=
=
(r − 1)
7
r
Javob:
')
(xg )
@
x
2
0.Muntazam ko’pburchak tomoni unga tashqi
chizilgan aylananing
36
°
li yoyni tortib turadi.
Muntazam ko’pburchakning tomonlari sonini
toping.
)12 3) ))6 )8
Yechim:
Muntazam ko’pburchakning markaziy burchagi
36
°
ga teng,
N =
T8z
°
|
.
36
°
=
360
°
J , J = 10.
Javob:
3)
21.ABC uchburchakda
D va
E nuqtalar
BC
tomonni
uchta
teng
qismlarga
bo’ladi
(
BD=DE=EC),
F va
G nuqtalar esa
AD kesmani
uchta teng qismlarga bo’ladi (AF=FG=GD).
AFE uchburchak yuzining
ABC uchburchak
yuziga nisbatini toping.
)
1
12 )
1
4 ))
1
3 ?) I
Yechim:
}
~•€
= }
}
•€•
= 2}
}
~€‚
= }
~•€
= 3}
2 ∙ }
~ĥ
= }
~•‚
= 6}
}
~ƒ‚
= }
~ĥ
+ }
~•‚
= 3} + 6} = 9}
}
~•€
}
~ƒ‚
=
}
9} =
1
9
Javob:
?)
I
22.ABCDEF muntazam oltiburchakda AC, CE,
BF,
FD diagonallar o’tkazilgan.
AC va
BF
diagonallar L nuqtada, CE va FD diagonallar K
nuqtada kesishadi. Agar oltiburchak tomoni
2√3
ga teng bo’lsa,
„)v… to’rtburchak yuzini toping.
)5√3 3)A√5 ))9√3 )6√3
Yechim:
„)v… to’rtburchak 60
°
li romb. Rombning
tomoni
7
√T
ga teng (
= 2√3).
}
†‚‡•
= 9
7
√T
:
7
OPJ60
°
= 9
7
√T
∙ 2√3:
7
∙
√T
7
=
16 ∙
√T
7
= 8√3
Javob:
3)A√5
23.ABCD trapetsiyaning yuzi 24 ga teng,
asoslari DC=6, AB=2. BC tomonidan E nuqta
olingan
bo’lib,
BE=2EC
bo’lsa,
ADE
uchburchak yuzini toping.
)12 )21 ) ˆ )16
Yechim:
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
7‰
T
va
‰
T
ga teng bo’ladi.
}
~ƒ‚•
=
(
+ ) ) ∙ ℎ
2
=
(2 + 6) ∙ ℎ
2
= 4ℎ = 24
ℎ = 6
}
~ƒ€
=
∙ 2ℎ
3
2
=
2 ∙ 2ℎ
6 =
2 ∙ 6
3 = 4
}
•€‚
=
) ∙ ℎ3
2 =
6 ∙ 6
6 = 6
}
~ƒ‚•
= }
~ƒ€
+ }
•€‚
+ }
~•€
24 = 6 + 4 + }
~•€
}
~•€
= 14
Javob:
) ˆ
24.
(3; 0) va (−1; 2) nuqtalardan o’tuvchi
hamda markazi
0 = + + 2 to’g’ri chiziqda
yotgan aylana tenglamasini toping.
)(+ − 3)
7
+ (0 − 5)
7
= 25
)(+ − 4)
7
+ (0 − 5)
7
= 25
))(+ − 3)
7
+ (0 − 4)
7
= 25
)(+ − 5)
7
+ (0 − 3)
7
= 25
Yechim:
(+ − )
7
+ (0 − )
7
= ‹
7
a (3 − )
7
+ (0 − )
7
= ‹
7
(−1 − )
7
+ (2 − )
7
= ‹
7
b
, (3 − )
7
+ (0 − )
7
= (−1 − )
7
+ (2 − )
7
,
9 − 6 +
7
+
7
= 1 + 2 +
7
+ 4 − 4 +
7
,
Œ = @• −
0 = + + 2 , Œ = • + @
Ž = 2 − 1
= + 2
b ,
= 3, = 5, ‹ = 5
(+ − 3)
7
+ (0 − 5)
7
= 25
Javob:
')(6 − 5)
@
+ (4 − \)
@
= @\
25.
= {1; 3; 5; 6; 8; 10} va
= {5; 6; 7; 8; 10}
to’plamlar berilgan.
∪ to’plam elementlari sonini
toping.
)8 )11 )‘ )6
Yechim:
Har
qanday
ikkita
to’plamning
barcha
elementlaridan, ularni takrorlamasdan tuzilgan
to’plamga shu to’plamning birlashmasi (yoki
yig’indisi) deyiladi.
∪ = {1; 3; 5; 6; 7; 8; 10}
Javob:
)‘
Testlar yechilishi davomida yo’l qo’yilgan
xatolar uchun uzr!