Đáp án:
$\begin{array}{l}
e)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} - 2} \right) = 15\\
\Rightarrow {x^3} + {2^3} - {x^3} + 2x = 15\\
\Rightarrow 2x + 8 = 15\\
\Rightarrow 2x = 7\\
\Rightarrow x = \dfrac{7}{2}\\
\text{Vậy}\,x = \dfrac{7}{2}\\
f)\left( {2x - 1} \right)\left( {2x + 3} \right) - {\left( {2x + 3} \right)^2} = 7\\
\Rightarrow \left( {2x + 3} \right)\left( {2x - 1 - 2x - 3} \right) = 7\\
\Rightarrow \left( {2x + 3} \right).\left( { - 4} \right) = 7\\
\Rightarrow 2x + 3 = - \dfrac{7}{4}\\
\Rightarrow 2x = \dfrac{{ - 7}}{4} - 3 = \dfrac{{ - 19}}{4}\\
\Rightarrow x = \dfrac{{ - 19}}{8}\\
\text{Vậy}\,x = \dfrac{{ - 19}}{8}\\
g)2x\left( {x + 5} \right) - x - 5 = 0\\
\Rightarrow \left( {x + 5} \right)\left( {2x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x + 5 = 0\\
2x - 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 5\\
x = \dfrac{1}{2}
\end{array} \right.\\
\text{Vậy}\,x = - 5;x = \dfrac{1}{2}\\
h){\left( {3x + 2} \right)^2} = 4{\left( {x - 3} \right)^2}\\
\Rightarrow {\left( {3x + 2} \right)^2} = {\left( {2x - 6} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
3x + 2 = 2x - 6\\
3x + 2 = - 2x + 6
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 8\\
5x = 4
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 8\\
x = \dfrac{4}{5}
\end{array} \right.\\
\text{Vậy}\,x = - 8;x = \dfrac{4}{5}
\end{array}$
