Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = 5
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ:
\(\left\{ \begin{array}{l}
{x^2} + 2x - 3 \ge 0\\
5 - x \ge 0\\
{x^2} + x + 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 1\\
x \le - 3
\end{array} \right.\\
x \le 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x \le - 3\\
1 \le x \le 5
\end{array} \right.\,\,\,\left( * \right)\)
Ta có:
\(\begin{array}{l}
\sqrt {{x^2} + 2x - 3} + \sqrt {5 - x} = \sqrt {{x^2} + x + 2} \\
\Leftrightarrow {x^2} + 2x - 3 + 2\sqrt {\left( {{x^2} + 2x - 3} \right)\left( {5 - x} \right)} + 5 - x = {x^2} + x + 2\\
\Leftrightarrow {x^2} + x + 2 + 2\sqrt {\left( {{x^2} + 2x - 3} \right)\left( {5 - x} \right)} = {x^2} + x + 2\\
\Leftrightarrow \sqrt {\left( {{x^2} + 2x - 3} \right)\left( {5 - x} \right)} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + 2x - 3 = 0\\
5 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - 3\\
x = 5
\end{array} \right.\left( {t/m} \right)
\end{array}\)
