Giải hệ phương trình của từng đáp án ta được:
+) Đáp án A:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 2x \le 0\\
2x + 1 < 3x + 2
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x\left( {x - 2} \right) \le 0\\
3x - 2x > 1 - 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
0 \le x \le 2\\
x > - 1
\end{array} \right. \Leftrightarrow 0 \le x \le 2.
\end{array}\)
+) Đáp án B:
\(\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - 4 > 0\\
\frac{1}{{x + 2}} < \frac{1}{{x + 1}}
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x + 2 \ne 0\\
x + 1 \ne 0\\
\left( {x - 2} \right)\left( {x + 2} \right) > 0\\
\frac{1}{{x + 1}} - \frac{1}{{x + 2}} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ne - 1\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
\frac{{x + 2 - x - 1}}{{\left( {x + 1} \right)\left( {x + 2} \right)}} > 0
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ne - 1\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
\frac{1}{{\left( {x + 1} \right)\left( {x + 2} \right)}} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ne - 1\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
\left( {x + 1} \right)\left( {x + 2} \right) > 0\;\;\left( {do\;\;1 > 0} \right)
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 2\\
x \ne - 1\\
\left[ \begin{array}{l}
x > 2\\
x < - 2
\end{array} \right.\\
\left[ \begin{array}{l}
x > - 1\\
x < - 2
\end{array} \right.
\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}
x > 2\\
x > - 2
\end{array} \right..
\end{array}\)
+) Đáp án C:
\(\left\{ \begin{array}{l}
{x^2} - 5x + 2 < 0\\
{x^2} + 8x + 1 \le 0
\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}
\frac{{5 - \sqrt {17} }}{2} < x < \frac{{5 + \sqrt {17} }}{2}\\
- 4 - \sqrt {17} < x < - 4 + \sqrt {17}
\end{array} \right.\\ \Leftrightarrow x \in \emptyset .\)
Chọn C.