Tìm x:
a) \(x + \dfrac{2}{5} = \dfrac{3}{{10}}\)
b) \(\dfrac{1}{6}x - 3 = \dfrac{{ - 2}}{3}\)
c) \({\left( {\dfrac{1}{5} - x} \right)^2} = \dfrac{{16}}{9}\)
a) \(x + \dfrac{2}{5} = \dfrac{3}{{10}}\)
\( \Leftrightarrow x = \dfrac{3}{{10}} - \dfrac{2}{5} = \dfrac{{3 - 2.2}}{{10}} = \dfrac{{ - 1}}{{10}}\)
b) \(\dfrac{1}{6}x - 3 = \dfrac{{ - 2}}{3}\)
\( \Leftrightarrow \dfrac{1}{6}x = \dfrac{{ - 2}}{3} + 3 = \dfrac{7}{3}\)
\( \Leftrightarrow x = \dfrac{7}{3}.6 = 14\)
c) \({\left( {\dfrac{1}{5} - x} \right)^2} = \dfrac{{16}}{9}\)
\( \Leftrightarrow \left| {\dfrac{1}{5} - x} \right| = \dfrac{4}{3}\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\dfrac{1}{5} - x = \dfrac{4}{3}}\\{\dfrac{1}{5} - x = {\rm{\;}} - \dfrac{4}{3}}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = \dfrac{1}{5} - \dfrac{4}{3}}\\{x = \dfrac{1}{5} + \dfrac{4}{3}}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = {\rm{\;}} - \dfrac{{17}}{{15}}}\\{x = \dfrac{{23}}{{15}}}\end{array}} \right..\)
\(\begin{array}{*{20}{l}}{d)\;\;\dfrac{{x - 1}}{6} = \dfrac{{x + 3}}{5}}\\{ \Leftrightarrow 5.\left( {x - 1} \right) = 6.\left( {x + 3} \right)}\\{ \Leftrightarrow 5x - 5 = 6x + 18}\\{ \Leftrightarrow 6x - 5x = {\rm{\;}} - 5 - 18}\\{ \Leftrightarrow x = {\rm{\;}} - 23}\end{array}\)
