Giải bài 4 trang 25 SGK Toán 7 tập 1 - Chân trời sáng tạo

Đề bài

Tìm x, biết:

a)$x + \frac{3}{5} = \frac{2}{3};$                      

b)$\frac{3}{7} - x = \frac{2}{5};$

c)$\frac{4}{9} - \frac{2}{3}x = \frac{1}{3};$                   

d)$\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}$

Lời giải chi tiết

a)

$\begin{array}{l}x + \frac{3}{5} = \frac{2}{3}\\x = \frac{2}{3} - \frac{3}{5}\\x = \frac{{10}}{{15}} - \frac{9}{{15}}\\x = \frac{1}{{15}}\end{array}$                      

Vậy $x = \frac{1}{{15}}$.

b)

$\begin{array}{l}\frac{3}{7} - x = \frac{2}{5}\\x = \frac{3}{7} - \frac{2}{5}\\x = \frac{{15}}{{35}} - \frac{{14}}{{35}}\\x = \frac{1}{{35}}\end{array}$

Vậy $x = \frac{1}{{35}}$.

c)

$\begin{array}{l}\frac{4}{9} - \frac{2}{3}x = \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{1}{3}\\\frac{2}{3}x = \frac{4}{9} - \frac{3}{9}\\\frac{2}{3}x = \frac{1}{9}\\x = \frac{1}{9}:\frac{2}{3}\\x = \frac{1}{9}.\frac{3}{2}\\x = \frac{1}{6}\end{array}$                   

Vậy $x = \frac{1}{6}$.

d)

$\begin{array}{l}\frac{3}{{10}}x - 1\frac{1}{2} = \left( {\frac{{ - 2}}{7}} \right):\frac{5}{{14}}\\\frac{3}{{10}}x - \frac{3}{2} = \left( {\frac{{ - 2}}{7}} \right).\frac{{14}}{5}\\\frac{3}{{10}}x - \frac{3}{2} = \frac{{ - 4}}{5}\\\frac{3}{{10}}x = \frac{{ - 4}}{5} + \frac{3}{2}\\\frac{3}{{10}}x = \frac{{ - 8}}{{10}} + \frac{{15}}{{10}}\\\frac{3}{{10}}x = \frac{7}{{10}}\\x = \frac{7}{{10}}:\frac{3}{{10}}\\x = \frac{7}{3}\end{array}$

Vậy $x = \frac{7}{3}$.