Đáp án:
$\begin{array}{l}
g)3x - 2\left| {x + 1} \right| = 5\\
\Rightarrow 2\left| {x + 1} \right| = 3x - 5\\
\Rightarrow \left[ \begin{array}{l}
2x + 2 = 3x - 5\\
2x + 2 = - 3x + 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 7\\
x = \dfrac{3}{5}
\end{array} \right.\\
h)\dfrac{{x - 2}}{{x + 3}} = \dfrac{2}{3}\left( {dk:x \ne - 3} \right)\\
\Rightarrow 3\left( {x - 2} \right) = 2\left( {x + 3} \right)\\
\Rightarrow 3x - 6 = 2x + 6\\
\Rightarrow x = 12\left( {tm} \right)\\
i)\dfrac{{x - 1}}{{24}} = \dfrac{6}{{x - 1}}\left( {dk:x \ne 1} \right)\\
\Rightarrow {\left( {x - 1} \right)^2} = 24.6 = 144\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 12\\
x - 1 = - 12
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 13\\
x = - 11
\end{array} \right.\left( {tm} \right)\\
k)\dfrac{{x - 2}}{{2016}} + \dfrac{{x - 3}}{{2015}} + \dfrac{{x - 4}}{{2014}} + \dfrac{{x - 5}}{{2013}} = 4\\
\Rightarrow \dfrac{{x - 2}}{{2016}} - 1 + \dfrac{{x - 3}}{{2015}} - 1 + \dfrac{{x - 4}}{{2014}} - 1 + \dfrac{{x - 5}}{{2013}} - 1 = 0\\
\Rightarrow \dfrac{{x - 2 - 2016}}{{2016}} + \dfrac{{x - 3 - 2015}}{{2015}} + \\
\dfrac{{x - 4 - 2014}}{{2014}} + \dfrac{{x - 5 - 2013}}{{2013}} = 0\\
\Rightarrow \dfrac{{x - 2018}}{{2016}} + \dfrac{{x - 2018}}{{2015}} + \dfrac{{x - 2018}}{{2014}} + \dfrac{{x - 2018}}{{2013}} = 0\\
\Rightarrow \left( {x - 2018} \right)\left( {\dfrac{1}{{2016}} + \dfrac{1}{{2015}} + \dfrac{1}{{2014}} + \dfrac{1}{{2013}}} \right) = 0\\
\Rightarrow x - 2018 = 0\\
\Rightarrow x = 2018\\
m)5\sqrt {x - 3} - 4 = 16\left( {dkxd:x \ge 3} \right)\\
\Rightarrow 5\sqrt {x - 3} = 16 + 4 = 20\\
\Rightarrow \sqrt {x - 3} = 4\\
\Rightarrow x - 3 = 16\\
\Rightarrow x = 19\left( {tm} \right)\\
n){\left( {x - 1} \right)^2} - 3\left( {x - 1} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {x - 1 - 3} \right) = 0\\
\Rightarrow \left( {x - 1} \right)\left( {x - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 4 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.
\end{array}$
