Đáp án:
$x \in \left( { - 1;2} \right) \cup \left( { - \infty ; - 4} \right)$
Giải thích các bước giải:
ĐKXĐ $x\ne -1;x\ne 2$
Ta có:
$f\left( x \right) = \dfrac{1}{{x + 1}} + \dfrac{2}{{2 - x}}$
Để $f\left( x \right) > 0$
$\begin{array}{l}
\Leftrightarrow \dfrac{1}{{x + 1}} + \dfrac{2}{{2 - x}} > 0\\
\Leftrightarrow \dfrac{{2 - x + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {2 - x} \right)}} > 0\\
\Leftrightarrow \dfrac{{x + 4}}{{\left( {x + 1} \right)\left( {2 - x} \right)}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 4 > 0\\
\left( {x + 1} \right)\left( {2 - x} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 4 < 0\\
\left( {x + 1} \right)\left( {2 - x} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 4\\
\left( {x + 1} \right)\left( {x - 2} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 4\\
\left( {x + 1} \right)\left( {x - 2} \right) > 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 4\\
- 1 < x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 4\\
\left[ \begin{array}{l}
x > 2\\
x < - 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- 1 < x < 2\\
x < - 4
\end{array} \right.\\
\Leftrightarrow x \in \left( { - 1;2} \right) \cup \left( { - \infty ; - 4} \right)
\end{array}$
Vậy $x \in \left( { - 1;2} \right) \cup \left( { - \infty ; - 4} \right)$ để $f\left( x \right) > 0$
