Đáp án:
$\begin{array}{l}
Dkxd:x \ne 0;x \ne - 1;x \ne 5;x \ne - 5\\
a)P = \left( {\dfrac{{15 - x}}{{{x^2} - 25}} + \dfrac{2}{{x + 5}}} \right):\dfrac{{x + 1}}{{2{x^2} - 10x}}\\
= \dfrac{{15 - x + 2\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}.\dfrac{{2x\left( {x - 5} \right)}}{{x + 1}}\\
= \dfrac{{15 - x + 2x - 10}}{{x + 5}}.\dfrac{{2x}}{{x + 1}}\\
= \dfrac{{x + 5}}{{x + 5}}.\dfrac{{2x}}{{x + 1}}\\
= \dfrac{{2x}}{{x + 1}}\\
b)\left| {2x - 3} \right| = 7\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 7\\
2x - 3 = - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 10\\
2x = - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\left( {ktm} \right)\\
x = - 2\left( {tm} \right)
\end{array} \right.\\
+ khi:x = - 2\\
\Leftrightarrow P = \dfrac{{2x}}{{x + 1}} = \dfrac{{2.\left( { - 2} \right)}}{{ - 2 + 1}} = 4\\
c)P = \dfrac{{2x}}{{x + 1}} = \dfrac{{2x + 2 - 2}}{{x + 1}}\\
= 2 - \dfrac{2}{{x + 1}} \in Z\\
\Leftrightarrow \dfrac{2}{{x + 1}} \in Z\\
\Leftrightarrow \left( {x + 1} \right) \in \left\{ { - 2; - 1;1;2} \right\}\\
\Leftrightarrow x \in \left\{ { - 3; - 2;0;1} \right\}\\
Do:x \ne 0\\
\Leftrightarrow x \in \left\{ { - 3; - 2;1} \right\}\\
Vay\,x \in \left\{ { - 3; - 2;1} \right\}
\end{array}$
