Đáp án:
$S = \left\{ 1 \right\}$
Giải thích các bước giải:
ĐKXĐ: $\left[ \begin{array}{l}
x \ge 1\\
x \le - 2
\end{array} \right.$
Ta có:
$\sqrt {{x^2} - 1} + \sqrt {{x^2} - x} \le \sqrt {{x^2} + x - 2} \left( 1 \right)$
$\begin{array}{l}
+ )TH1:x \ge 1\\
\left( 1 \right) \Leftrightarrow \sqrt {x - 1} .\sqrt {x + 1} + \sqrt {x - 1} .\sqrt x \le \sqrt {x - 1} .\sqrt {x + 2} \\
\Leftrightarrow \sqrt {x - 1} \left( {\sqrt {x + 1} + \sqrt x } \right) \le \sqrt {x - 1} .\sqrt {x + 2} \\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 1} = 0\\
\sqrt {x + 1} + \sqrt x \le \sqrt {x + 2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
2x + 1 + 2\sqrt {x\left( {x + 1} \right)} \le x + 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
2\sqrt {x\left( {x + 1} \right)} \le 1 - x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
2\sqrt {x\left( {x + 1} \right)} = 1 - x = 0\left( {do:1 - x \le 0} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left\{ \begin{array}{l}
x = 1\\
\left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.
\end{array} \right.\left( l \right)
\end{array} \right.\\
\Leftrightarrow x = 1
\end{array}$
$\begin{array}{l}
+ )TH2:x \le - 2\\
\left( 1 \right) \Leftrightarrow \sqrt {1 - x} .\sqrt { - x - 1} + \sqrt {1 - x} .\sqrt { - x} \le \sqrt {1 - x} .\sqrt { - 2 - x} \\
\Leftrightarrow \sqrt {1 - x} \left( {\sqrt { - x - 1} + \sqrt { - x} } \right) \le \sqrt {1 - x} .\sqrt { - 2 - x} \\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {1 - x} = 0\\
\sqrt { - x - 1} + \sqrt { - x} \le \sqrt { - x - 2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( l \right)\\
- 2x - 1 + 2\sqrt {x\left( {x + 1} \right)} \le - x - 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( l \right)\\
2\sqrt {x\left( {x + 1} \right)} \le x - 1\left( {vn,do:x \le - 2 \Rightarrow x - 1 \le - 3 < 0} \right)
\end{array} \right.
\end{array}$
Vậy bất phương trình có tập nghiệm $S = \left\{ 1 \right\}$
