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Câu hỏi: Tìm tập xác định của hàm số $y=\sqrt{\left( {{x}^{2}}-x-2 \right)\ln \left( x+2 \right)}$
A. {-1} $\cup $ [2; + $\infty $ )
B. [-2; + $\infty $ )
C. [-2;-1] $\cup $ [2; + $\infty $ )
D. [2; + $\infty $ )
A. {-1} $\cup $ [2; + $\infty $ )
B. [-2; + $\infty $ )
C. [-2;-1] $\cup $ [2; + $\infty $ )
D. [2; + $\infty $ )
Điều kiện xác định: $\left\{ \begin{aligned}
& x+2>0 \\
& \left( {{x}^{2}}-x-2 \right)\ln \left( x+2 \right)\ge 0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x>-2 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{x}^{2}}-x-2\ge 0 \\
& x+2\ge 1 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{x}^{2}}-x-2\le 0 \\
& x+2\le 1 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x>-2 \\
& \left[ \begin{aligned}
& x\ge 2 \\
& x\le -1 \\
\end{aligned} \right. \\
& x\ge -1 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& x>-2 \\
& -1\le x\le 2 \\
& x\le -1 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x\ge 2 \\
& x=-1 \\
\end{aligned} \right.$
& x+2>0 \\
& \left( {{x}^{2}}-x-2 \right)\ln \left( x+2 \right)\ge 0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& x>-2 \\
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {{x}^{2}}-x-2\ge 0 \\
& x+2\ge 1 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {{x}^{2}}-x-2\le 0 \\
& x+2\le 1 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& x>-2 \\
& \left[ \begin{aligned}
& x\ge 2 \\
& x\le -1 \\
\end{aligned} \right. \\
& x\ge -1 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& x>-2 \\
& -1\le x\le 2 \\
& x\le -1 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x\ge 2 \\
& x=-1 \\
\end{aligned} \right.$
Đáp án A.