Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
h\left( x \right) = \frac{{\left( {x + 1} \right)\left( {4 - x} \right)}}{{1 - 2x}} = \frac{{\left( {x + 1} \right).\left( {x - 4} \right)}}{{2x - 1}}\,\,\,\,\,\left( {x \ne \frac{1}{2}} \right)\\
TH1:\,\,x \le - 1 \Rightarrow \left\{ \begin{array}{l}
x + 1 \le 0\\
x - 4 < 0\\
2x - 1 < 0
\end{array} \right. \Rightarrow h\left( x \right) \le 0\\
TH2:\,\, - 1 < x < \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}
x + 1 > 0\\
x - 4 < 0\\
2x - 1 < 0
\end{array} \right. \Rightarrow h\left( x \right) > 0\\
TH3:\,\,\,\,\frac{1}{2} < x < 4 \Rightarrow \left\{ \begin{array}{l}
x + 1 > 0\\
x - 4 < 0\\
2x - 1 > 0
\end{array} \right. \Rightarrow h\left( x \right) < 0\\
TH4:\,\,x \ge 4 \Rightarrow \left\{ \begin{array}{l}
x + 1 > 0\\
x - 4 \ge 0\\
2x - 1 > 0
\end{array} \right. \Rightarrow h\left( x \right) \ge 0
\end{array}\)
