Đáp án:
$\begin{array}{l}
Dkxd:x\# 1;x\# - 1;x\# 0\\
a)A = \dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}}\\
= \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{4x}}{{{x^2} - 1}}\\
b)P = A:B\\
= \dfrac{{4x}}{{{x^2} - 1}}:\dfrac{{2x}}{{5x - 5}}\\
= \dfrac{{4x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{5\left( {x - 1} \right)}}{{2x}}\\
= \dfrac{{10}}{{x + 1}}\\
c)x = - \dfrac{3}{4}\left( {tmdk} \right)\\
P = \dfrac{{10}}{{\dfrac{{ - 3}}{4} + 1}} = 40\\
d)P = 2\\
\Leftrightarrow \dfrac{{10}}{{x + 1}} = 2\\
\Leftrightarrow x + 1 = 5\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
e)P \in Z\\
\Leftrightarrow 10 \vdots \left( {x + 1} \right)\\
\Leftrightarrow \left( {x + 1} \right) \in \left\{ { - 10; - 5; - 2; - 1;1;2;5;10} \right\}\\
\Leftrightarrow x \in \left\{ { - 11; - 6; - 3; - 2;0;1;4;9} \right\}\\
Do:x\# 1;x\# 0;x\# - 1\\
\Leftrightarrow x \in \left\{ { - 11; - 6; - 3; - 2;4;9} \right\}
\end{array}$
