Bài 4 trang 18 SBT Toán lớp 7 tập 1 Chân trời sáng tạo: Tìm x, biết:...

Tìm x, biết:

a) $x + \dfrac{3}{7} = \dfrac{2}{5}$

b) $\dfrac{3}{2} - x = \dfrac{4}{5}$

c) $\dfrac{5}{9} - \dfrac{1}{3}x = \dfrac{2}{3}$

d) $\dfrac{3}{5}x - 1\dfrac{1}{5} = \dfrac{{ - 3}}{{14}}:\dfrac{5}{7}$

Lý thuyết Quy tắc dấu ngoặc và quy tắc chuyển vế SGK Toán 7 Chân trời sáng tạo (baitapsgk.com)

$\begin{array}{l}a)x + \dfrac{3}{7} = \dfrac{2}{5}\\ \Leftrightarrow x = \dfrac{2}{5} - \dfrac{3}{7}\end{array}$

$\begin{array}{l} \Leftrightarrow x = \dfrac{{14}}{{35}} - \dfrac{{15}}{{35}}\\ \Leftrightarrow x = \dfrac{{ - 1}}{{35}}\end{array}$

Vậy $x = \dfrac{{ - 1}}{{35}}$

$\begin{array}{l}b)\dfrac{3}{2} - x = \dfrac{4}{5}\\ \Leftrightarrow x = \dfrac{3}{2} - \dfrac{4}{5}\\ \Leftrightarrow x = \dfrac{{15}}{{10}} - \dfrac{8}{{10}}\\ \Leftrightarrow x = \dfrac{7}{{10}}\end{array}$

Vậy $x = \dfrac{{ 7}}{{10}}$

$\begin{array}{l}c)\dfrac{5}{9} - \dfrac{1}{3}x = \dfrac{2}{3}\\ \Leftrightarrow \dfrac{1}{3}x = \dfrac{5}{9} - \dfrac{2}{3}\\ \Leftrightarrow \dfrac{1}{3}x = \dfrac{{ - 1}}{9}\\ \Leftrightarrow x = \dfrac{{ - 1}}{9}:\dfrac{1}{3}\\ \Leftrightarrow x = \dfrac{{ - 1}}{3}\end{array}$

Vậy $x = \dfrac{{ - 1}}{{3}}$

$\begin{array}{l}d)\dfrac{3}{5}x - 1\dfrac{1}{5} = \dfrac{{ - 3}}{{14}}:\dfrac{5}{7}\\ \Leftrightarrow \dfrac{3}{5}x - \dfrac{6}{5} = \dfrac{{ - 3}}{{10}}\\ \Leftrightarrow \dfrac{3}{5}x = \dfrac{{ - 3}}{{10}} + \dfrac{6}{5}\\ \Leftrightarrow \dfrac{3}{5}x = \dfrac{9}{{10}}\\ \Leftrightarrow x = \dfrac{9}{{10}}:\dfrac{3}{5}\\ \Leftrightarrow x = \dfrac{3}{2}\end{array}$

Vậy $x = \dfrac{{ 3}}{{2}}$