Bài 4 trang 18

Đề bài

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

Lời giải chi tiết

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

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

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 \(x = \dfrac{{ 3}}{{2}}\)