Đáp án: $x = \frac{{3 + 3\sqrt 5 }}{2}$
Giải thích các bước giải:
$\begin{array}{l}
ĐKxđ: - 2 \le x \le 5\\
\sqrt {x + 2} + \sqrt {5 - x} = a\left( {a > 0} \right)\\
\Rightarrow {a^2} = x + 2 + 5 - x + 2\sqrt {x + 2} .\sqrt {5 - x} \\
\Rightarrow {a^2} = 7 + 2\sqrt {\left( {x - 2} \right)\left( {5 - x} \right)} \\
\Rightarrow \sqrt {\left( {x - 2} \right)\left( {5 - x} \right)} = \frac{{{a^2} - 7}}{2}\\
\Rightarrow pt:a + \frac{{{a^2} - 7}}{2} = 4\\
\Rightarrow {a^2} + 2a - 7 - 8 = 0\\
\Rightarrow {a^2} + 2a - 15 = 0\\
\Rightarrow \left( {a - 3} \right)\left( {a + 5} \right) = 0\\
\Rightarrow a = 3\left( {do\,:a > 0} \right)\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {x + 2} + \sqrt {5 - x} = 3\\
\sqrt {\left( {x - 2} \right)\left( {5 - x} \right)} = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\sqrt {x + 2} = \frac{{3 - \sqrt 5 }}{2}\\
\sqrt {5 - x} = \frac{{3 + \sqrt 5 }}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
\sqrt {x + 2} = \frac{{3 + \sqrt 5 }}{2}\\
\sqrt {5 - x} = \frac{{3 - \sqrt 5 }}{2}
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 = \frac{{7 - 3\sqrt 5 }}{2}\\
5 - x = \frac{{7 + 3\sqrt 5 }}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 = \frac{{7 + 3\sqrt 5 }}{2}\\
5 - x = \frac{{7 - 3\sqrt 5 }}{2}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = \frac{{1 - 3\sqrt 5 }}{2}\\
x = \frac{{3 - 3\sqrt 5 }}{2}
\end{array} \right.\left( {ktm} \right)\\
\left\{ \begin{array}{l}
x = \frac{{3 + 3\sqrt 5 }}{2}\\
x = \frac{{3 + 3\sqrt 5 }}{2}
\end{array} \right.
\end{array} \right. \Rightarrow x = \frac{{3 + 3\sqrt 5 }}{2}\left( {tm} \right)
\end{array}$
