Đáp án:
$\begin{array}{l}
70.a)Dkxd:\left\{ \begin{array}{l}
{x^2} + 2x - 8 \ne 0\\
{x^2} + 2x - 3 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {x - 2} \right)\left( {x + 4} \right) \ne 0\\
\left( {x + 3} \right)\left( {x - 1} \right) \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 4\\
x \ne - 3\\
x \ne 2\\
x \ne 1
\end{array} \right.\\
Đặt\,{x^2} + 2x - 8 = a\left( {a \ne 0} \right)\\
\Rightarrow {x^2} + 2x - 3 = a + 5\\
\Rightarrow \frac{{24}}{a} - \frac{{15}}{{a + 5}} = 2\\
\Rightarrow \frac{{24\left( {a + 5} \right) - 15a}}{{a\left( {a + 5} \right)}} = 2\\
\Rightarrow 24a + 120 - 15a = 2\left( {{a^2} + 5a} \right)\\
\Rightarrow 9a + 120 = 2{a^2} + 10a\\
\Rightarrow 2{a^2} - a - 120 = 0\\
\Rightarrow \left( {2a + 15} \right)\left( {a - 8} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
a = - \frac{{15}}{2} = {x^2} + 2x - 8\\
a = 8 = {x^2} + 2x - 8
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x^2} + 2x - \frac{1}{2} = 0\\
{x^2} + 2x - 16 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{{ - 2 \pm \sqrt 6 }}{2}\\
x = - 1 \pm \sqrt {17}
\end{array} \right.\left( {tm} \right)
\end{array}$
