Đáp án:
\(\begin{array}{l}
d) - \dfrac{2}{{x + 1}}\\
e)x > - 1\\
f)\left[ \begin{array}{l}
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
d)DK:x \ne \left\{ { - 1;1;3} \right\}\\
P = \dfrac{{x + 3 - 3\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}:\dfrac{{x - 1 - 2}}{{x - 1}}\\
= \dfrac{{ - 2x + 6}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}:\dfrac{{x - 3}}{{x - 1}}\\
= \dfrac{{ - 2\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x - 1}}{{x - 3}}\\
= - \dfrac{2}{{x + 1}}\\
e)P < 0\\
\to - \dfrac{2}{{x + 1}} < 0\\
\to x + 1 > 0\\
\to x > - 1\\
f)Q = x.P = x. - \dfrac{2}{{x + 1}}\\
= - \dfrac{{2x}}{{x + 1}} = - \dfrac{{2\left( {x + 1} \right) - 2}}{{x + 1}}\\
= - 2 + \dfrac{2}{{x + 1}}\\
Q \in Z\\
\to \dfrac{2}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
