Đáp án:
e. x=-2
Giải thích các bước giải:
\(\begin{array}{l}
a.x + \dfrac{5}{3} = \dfrac{7}{4} - \left( { - \dfrac{{12}}{5}} \right)\\
\to x + \dfrac{5}{3} = \dfrac{{83}}{{20}}\\
\to x = \dfrac{{149}}{{60}}\\
b.DK:x \ne 0\\
\dfrac{7}{{4x}} = \dfrac{5}{6} - \dfrac{2}{3}\\
\to \dfrac{7}{{4x}} = \dfrac{1}{6}\\
\to 4x = 42\\
\to x = \dfrac{{21}}{2}\\
c.x - \dfrac{1}{3} = \dfrac{7}{{35}}:\left( { - \dfrac{2}{{25}}} \right)\\
\to x - \dfrac{1}{3} = - \dfrac{5}{2}\\
\to x = - \dfrac{5}{2} + \dfrac{1}{3}\\
\to x = - \dfrac{{13}}{6}\\
d.\left[ \begin{array}{l}
x = 0\\
x - \dfrac{1}{3} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{3}
\end{array} \right.\\
e.\left( {\dfrac{3}{7} - \dfrac{2}{3}} \right)x = \dfrac{{10}}{{21}}\\
\to - \dfrac{5}{{21}}x = \dfrac{{10}}{{21}}\\
\to - 5x = 10\\
\to x = - 2
\end{array}\)
