Matematikadan

1) - 2,8 • a ■ (- 1,5)6 • 2c;
2) (2^-: 1-^-1 (-a) • 3,56 ■ (-2,l)c;
3 ) -7а|-з||б-(-4,8)с.
147
4 )—3x-(2,7)rl2,l2|
5) x • 1,8у • (- 5)z;
6) f 1 2 4—: 1—
2 8
u-2,5i>-(-l,4)u;.
493. 0 ‘xshash hadlarni ixchamlang:
1) 5a - 3a - 8 a;
2 )-5 ,1 b -3 ,1 6 + 4,26;
3 2
3) —9—x —1—x + 7,3x;
5 5
4) - 18,27a + 16,27a- 3,5a;
5) 8,36 + 1,76-2,8-6;
6)3—x + 2—x —5—x.
4 3 3
494. Qavslarni oching va oxshash hadlarni
chamlang:
1) 3a - 5 • ( 7 - 2a);
2) — 6 a + 3 ■ (2a + 1);
3) 4,1 • ( - 2,5a + 1,8) + 3,2a;
4 ) -8 - (4,56-2,7) + 37,76;
5) - 9 - 6 + 156-2-6;
7 7 • '
6) - 5 -^ 5 + 3 86- 4 ^ 5;
11 11
7) - 7 ^ - 2 , 5y + 9Ѓ}y,
О О
8 ) 5—x — 4,8x —10—x.
7 7
148
495. Qavslarni oching va oxshash hadlarni ixchamlang:
1) — 4 • (2 - 3x) — 5 • (- 3x + 1);
2) 7 • (5x - 3) - 2 • (15x- 8);
3) (5a - 36) • (-4) - 3 • (a - 6);
4) (2a - 6) • (-3) + 2 • (4a - 36);
5) - l,5(2a - 5,4) - 8 • (1,5 - 2,5a)
6 ) 3,2 • (5 — y) + 4,5 (2y — 5);
f i 4 ' 1 f-. 3 7 N
z - 1 — + 3 - 1— z +
I 2 1 , 7 I 11 И ,
8 ) (7,3 - 2 z ) •(—5 ) - 7,2 5 z + -
12
496. Ixtiyoriy ikkita ratsional son a va 6 uchun
quyidagi tengliklar to‘g‘ri bolishini isbotlang:
1) (a + 6) - (a - 6) = 26
2 ) (a + 6) + (a — 6) = 2 a
04 a + 6 a - 6 3 ) + ------ = a;
2 2
4) a + b
497. Yuqoridagi masalaning 3) va 4) bandlaridan
foydalanib, ushbu masalalarni hal qiling:
1) Ikki sonning yig‘indisi 117 ga, ayirmasi esa
27 ga teng. Shu sonlarni toping.
2) Ikki sonning yiglndisi 162 ga, ayirmasi esa
4 ga teng. Shu sonlarni toping.
3) Ikki sonning yiglndisi (-10) ga teng. Katta
sondan kichigini ayirilsa, 84 hosil boladi. Shu
sonlarni toping.
4) Ikki sonning yiglndisi 0 ga teng. Sonlarning
kichigidan kattasi ayirilsa, (- 5,5) chiqadi.
Shu sonlarni toping.
149
498. Tenglamani yeching:
1) 3,8-(-5x)-2,8=7,6*ll,2;
2) (—2,4)-3,5x-4,2=2,64-12,6;
3) 7 _ .
3
4)
_ 5 \ л 1
1— X •1 -
2 2 5
9 ^ _ 2 — ■X •8 -
14 J 3
1
3
5) 16,8 : я: = 11,4 : 15,2;
6) (- x ) : 1,3 = 3,9 : (-2,6);
7) 1,44 : (— 1,2) = 9,6 : (—x);
8 )-9,1 : 1,3- x : (-10,01).
Tenglamalarni yechish
499. 1) Tenglama deganda nimani tushunasiz?
Tenglamani yechish degandachi? Misollar
keltiring.
. 2 ) Tenglamaning ildizi nima? Tenglama ildizga
ega bolishi shartmi? Misollar keltiring.
3) Tenglama to‘g‘ri yechilganini qanday tekshirish
mumkin? Misollar keltiring.
4) Tenglamaning asosiy xossalarini ayting va
misollarda tushuntiring.
5) Chiziqli tenglamalarni yechishning al-
Xorazmiy uslubi nimadan iborat? Misollarda
izohlab bering. (Darsligingizdagi ≪Tarixiy
malumotlar≫ga qarang).
500. 1) —4, —3, —1, 0,1, — sonlardan qaysi biri ush-
3
150
bu tanglamalarning ildizlari boladi:
1) 5x + 3 — 4x + 15 : 5;
2) 2(x + 1) + 7 = 3x + 10;
3) 3(x - 2) = 8x + 1;
4) 6 x + 5 = 7 — - x;
3
5)4x+17 = 5x + 21;
6 ) 2x + 9 = 7x + 24.
2 ) Noma’lum x qatnashgan hadlarni tenglamaning
chap qismiga, malum (ozod) hadlarini
esa o‘ng qismiga o‘tkazib, ifodani soddalashtiring
va hosil bolgan tenglamani yeching:
1) 15x + 7 = 3x + 19;
2) 16 - 7x = 19 - lOx;
3) 2,5x - 8 = 12 - 2,5x;
4) 3,7x— 1,8=5,2 — 3,3x;
5) 15,4x - 4,8 = 5,4x + 5,2;
6 ) 3—x -1,8 = 9,3-2—x;
7) l L 2 ± x = 3 ^ x;
' 9 14 9 14
8 ) l,25x + 7— = -0,75x + 5—;
3 3
9) — 8,4 — 7,5x = 12,5x + 11,6;
1Г>ч_3 _ 5 , 5 , , 5
10)7—x - 2— = - 1—x - 1 1—.
8 7 8 7
501. Tenglamani yeching:
1) (5x + 6 ) - (3x — 4) = (x - 3) - (2x— 4);
2) (5 - 3x) - (7 - 2x)=(x - 4) - (3x - 1);
3) (Q,25x - 1,8) + (l,7x + 2,8) - (4 - 0,05x) = 2;
151
-5- (4 1 3^ 3 f l Л " ) —x - 1— + — —x + 1—
8 U 5j 4 U 3 J = 5 - Зх.
502. Tenglamani yeching:
5x -7
1)
3)
5)
7)
x + 4
2x + 3
5 -x
x + 3
x + 1
x - 1
x + 2
= 3;
_3
~ 5 ’
5.
" 3’
1
V
x + 1 1
x + 2 2
-3x + 4 _ 7
4x -1 ~ - 5 :
x + 1 _ 5.
x - 3 ~ 4 ’
- x + 7 11
1 — x ~ 5 ’
x ~h 7 2
Namuna:------= — tenglamani yeching.
4 - x 9
Yechish: Bu tenglamani proporsiya, ya’ni
ikki nisbatning tengligi deb qarash mumkin:
(x + 7) : (4 — x) = 2 : 9. Proporsiyaning asosiy
xossasiga muvofiq: 9(x + 7) = 2(4 — x), bundan,
qavslarni ochib 9x + 63 = 8 — 2x tenglamaga
kelamiz. Uni yechamiz: 9x + 2x = 8 - 63;
l lx = -55, x= - 55 : 11, x = -5 .
Tekshirish: -5 + 7 2 2
4 - (—5) 4 + 5 9
(tenglama-
2 2
ning chap qismi), demak, — = —. Javob: x = — 5.
9 9
503. Tenglamani yeching:
1) - 2,3 x - 0,2x + 1,07 x — — 1,573;
2) 6 • (1,75*+ 0,625) =-6,75;
3) — 0,94* + l,5x — 0,7x = — 2,24;
152
4) 0,97х — 0,7л; + 0,9х - 0,5х = - 12,73.
Masalalarni tenglama tuzib yeching.
Ularni arifmetik usulda hamyechib koring va
masala yechishning bu ikkala usulini taqqoslang.
504. 1) Beshta ketma-ket toq sonning yig‘indisi 375
ga teng. Shu sonlarni toping.
2) Beshta ketma-ket juft sonning yigmdisi
630 ga teng. Shu sonlarni toping.
505. Beshta sonning o‘rta arifmetik qiymati (—1,6)
ga teng. Shu beshta songa yana bir x son
qo‘shib, o‘rta arifmetik qiymat hisoblangan
edi, u 1) 1 ,2 ga; 2 ) — ga teng chiqdi. x ni toping.
506. (Qadimgi masala). Kulol 25 ta ko‘zani bozorga
olib borish uchun arava yolladi. Bozorga olib
kelingan har bir butun ko‘za uchun kulol aravakashga
1500 so‘m beradi. Sindirib qoyilgan
har bir ko‘za uchun'aravakash kulolga 5000
so‘m tolaydi. Yo'lda bir nechta ko‘za sinib
qolipti va kulol aravakashga 18000 so‘m berdi.
Nechta ko‘za bozorga sinmay kelgan?
507. 1) Ikkita sonning biri ikkinchisidan 11 ta ortiq.
Katta sonning 30% i kichik sonning 40 %
idan 0 ,8 ga ko‘p. Shu sonlarni toping.
2) Ikkita sonning biri ikkinchisidan 18 ta kam.
Kichik sonning 30 % i katta sonning 20 % idan
0,9 ga ko‘p. Shu sonlarni toping.
153
508. 1) Beshta ketma-ket butun sonlar yiglndisi 0
ga teng. Shu sonlarni toping.
2) Beshta ketma-ket natural sonlar yig‘indisi
515 ga teng. Shu sonlarni toping.
509. 192 m masofada aravaning keyingi g‘ildiragi
oldingisiga qaraganda 2 0 ta kam aylanadi.
Oldingi glldirak aylanasi 3,2 m bolsa, keyingi
glldirak aylanasi uzunligini toping.
510. Charxpalakning chelakchalari bir-biridan bir
xil masofada joylashgan bolib, yonma-yon
turgan ikkita chelakcha orasidagi masofa
78,5 sm. Charxpalakning diametri 4 m bolsa,
bu charxpalakka nechta chelakcha osilgan?
(71=3,14 deb oling).
511. 1) Elektr quvvati bilan ishlaydigan doirasimon
arra tishlarining uchlari orasidagi aylana
bo‘ylab hisoblangan masofalar bir xil bolib,
2 ,1 sm ga teng. Arraning 72 ta tishi bolsa,
uning diametrini yuzdan birgacha aniqlikda
toping.