Đáp án:
$\begin{array}{l}
a)x - \dfrac{{5x + 2}}{6} = \dfrac{{7 - 3x}}{4}\\
\Leftrightarrow \dfrac{{12x - 10x - 4}}{{12}} = \dfrac{{21 - 9x}}{{12}}\\
\Leftrightarrow 2x - 4 = 21 - 9x\\
\Leftrightarrow 11x = 25\\
\Leftrightarrow x = \dfrac{{25}}{{11}}\\
Vậy\,x = \dfrac{{25}}{{11}}\\
b)\left| {x - 5} \right| + \left| {25 - {x^2}} \right| = 0\\
\Leftrightarrow \left| {x - 5} \right| + \left| {x - 5} \right|.\left| {x + 5} \right| = 0\\
\Leftrightarrow \left| {x - 5} \right|.\left( {\left| {x + 5} \right| + 1} \right) = 0\\
\Leftrightarrow \left| {x - 5} \right| = 0\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
c)\left| {x + 1} \right| - \left| {3 - 2x} \right| = 0\\
\Leftrightarrow \left| {x + 1} \right| = \left| {3 - 2x} \right|\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 3 - 2x\\
x + 1 = 2x - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 4
\end{array} \right.\\
Vậy\,x = \dfrac{2}{3};x = 4
\end{array}$
