Đáp án:
\(\begin{array}{l}
a)x \ne \pm 1\\
b)\dfrac{{2x}}{{x + 1}}\\
c)\left[ \begin{array}{l}
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 1\\
b)P = \dfrac{{2{x^2} + x\left( {x - 1} \right) - x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{2{x^2} + {x^2} - x - {x^2} - x}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{2{x^2} - 2x}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{2x}}{{x + 1}}\\
c)P = \dfrac{{2x}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) - 2}}{{x + 1}}\\
= 2 - \dfrac{2}{{x + 1}}\\
P \in Z \to \dfrac{2}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
