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) \(\dfrac{3}{5}x – 1\dfrac{1}{5} = \dfrac{{ – 3}}{{14}}:\dfrac{5}{7}\)
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\(\begin{array}{l}a)x + \dfrac{3}{7} = \dfrac{2}{5}\\ \Leftrightarrow x = \dfrac{2}{5} – \dfrac{3}{7}\end{array}\)
\(\begin{array}{l} \Leftrightarrow x = \dfrac{{14}}{{35}} – \dfrac{{15}}{{35}}\\ \Leftrightarrow x = \dfrac{{ – 1}}{{35}}\end{array}\)
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 \(x = \dfrac{{ 3}}{{2}}\)