Đáp án: $ - 1 \le x \le 0$
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {x + 1} - \sqrt {1 - x} \ge x\left( {dkxd: - 1 \le x \le 1} \right)\\
\Rightarrow \frac{{\left( {\sqrt {x + 1} - \sqrt {1 - x} } \right)\left( {\sqrt {x + 1} + \sqrt {1 - x} } \right)}}{{\sqrt {x + 1} + \sqrt {1 - x} }} \ge x\\
\Rightarrow \frac{{x + 1 - \left( {1 - x} \right)}}{{\sqrt {x + 1} + \sqrt {1 - x} }} \ge x\\
\Rightarrow \frac{{2x}}{{\sqrt {x + 1} + \sqrt {1 - x} }} - x \ge 0\\
\Rightarrow x.\left( {\frac{1}{{\sqrt {x + 1} + \sqrt {1 - x} }} - 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
\frac{1}{{\sqrt {x + 1} + \sqrt {1 - x} }} \ge 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
\frac{1}{{\sqrt {x + 1} + \sqrt {1 - x} }} \le 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
\sqrt {x + 1} + \sqrt {1 - x} \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
\sqrt {x + 1} + \sqrt {1 - x} \ge 0
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
2 + 2\sqrt {1 - {x^2}} \le 0\left( {vn} \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
2 + 2\sqrt {1 - {x^2}} \ge 0\left( {luôn\,đúng} \right)
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x \le 0\\
- 1 \le x \le 1
\end{array} \right.\\
\Rightarrow - 1 \le x \le 0
\end{array}$
