Đáp án: $x = 2$
Giải thích các bước giải:
$\begin{array}{l}
DKxd:\left\{ \begin{array}{l}
2 + x \ge 0\\
6 - x \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 2\\
x \le 6
\end{array} \right. \Leftrightarrow - 2 \le x \le 6\\
Dat:\sqrt {2 + x} + \sqrt {6 - x} = a\left( {a \ge 0} \right)\\
\Leftrightarrow 2 + x + 2\sqrt {\left( {2 + x} \right)\left( {6 - x} \right)} + 6 - x = {a^2}\\
\Leftrightarrow \sqrt {\left( {2 + x} \right)\left( {6 - x} \right)} = \frac{{{a^2} - 8}}{2}\\
Pt:a + \frac{{{a^2} - 8}}{2} = 8\\
\Leftrightarrow {a^2} + 2a - 8 = 16\\
\Leftrightarrow {a^2} + 2a - 24 = 0\\
\Leftrightarrow \left( {a - 4} \right)\left( {a + 6} \right) = 0\\
\Leftrightarrow a = 4\left( {do:a \ge 0} \right)\\
\Leftrightarrow \sqrt {\left( {2 + x} \right)\left( {6 - x} \right)} = \frac{{{a^2} - 8}}{2} = 4\\
\Leftrightarrow \left( {2 + x} \right)\left( {6 - x} \right) = 16\\
\Leftrightarrow - {x^2} + 4x + 12 = 16\\
\Leftrightarrow {x^2} - 4x + 4 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 0\\
\Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,x = 2
\end{array}$
