Tìm x, biết: a) –3x + 7 = 12 – 125; b) 1/3:(2x-1)=-4/21; c) [124 – (20 – 4x)] : 20 = 12

Giải SBT Toán 6 Cánh diều Bài tập cuối chương 5  

Bài 124 trang 60 SBT Toán 6 Tập 2:

Tìm x, biết:

a) –3x + 7 = 12 – 125;

b) $\frac{1}{3}:\left(2x-1\right)=\frac{-4}{21}$;

c) [124 – (20 – 4x)] : 20 = 12;

d) $\left(\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{8.9}+\frac{1}{9.10}\right).x=\frac{1}{5}$.

Lời giải

a) –3x + 7 = 12 – 125

–3x + 7 = –113

–3x = –113 – 7

–3x = –120

x = (–120) : (–3) = 40.

Vậy x = 40.

b) $\frac{1}{3}:\left(2x-1\right)=\frac{-4}{21}$

$2x-1=\frac{1}{3}:\left(\frac{-4}{21}\right)$

$2x-1=\frac{1}{3}.\frac{-21}{4}=\frac{-7}{4}$

$2x=\frac{-7}{4}+1=\frac{-3}{4}$

$x=\frac{-3}{4}:2=\frac{-3}{8}$.

Vậy $x=\frac{-3}{8}$.

c) [124 – (20 – 4x)] : 20 = 12

124 – (20 – 4x) = 12 . 20 

124 – (20 – 4x) = 240

20 – 4x = 124 – 240

20 – 4x = –116

4x = 20 – (–116)

4x = 136

x = 136 : 4

x = 34.

Vậy x = 34.

d) $\left(\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{8.9}+\frac{1}{9.10}\right).x=\frac{1}{5}$

$\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\mathrm{...}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right).x=\frac{1}{5}$

$\left(\frac{1}{2}-\frac{1}{10}\right).x=\frac{1}{5}$

$\left(\frac{5}{10}-\frac{1}{10}\right).x=\frac{1}{5}$

$\frac{4}{10}.x=\frac{1}{5}$

$\frac{2}{5}.x=\frac{1}{5}$

$x=\frac{1}{5}:\frac{2}{5}$

$x=\frac{1}{5}.\frac{5}{2}=\frac{1}{2}$

Vậy $x=\frac{1}{2}$.