Đáp án:
$\begin{array}{l}
y = f\left( x \right) = \sqrt {2 - x} + \sqrt {x + 2} \\
a)Dkxd:\left\{ \begin{array}{l}
2 - x \ge 0\\
x + 2 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 2\\
x \ge - 2
\end{array} \right.\\
\Leftrightarrow - 2 \le x \le 2\\
Vậy\,D = \left[ { - 2;2} \right]\\
b)f\left( a \right) = \sqrt {2 - a} + \sqrt {a + 2} \\
f\left( { - a} \right) = \sqrt {2 - \left( { - a} \right)} + \sqrt { - a + 2} \\
= \sqrt {2 - a} + \sqrt {a + 2} \\
\Leftrightarrow f\left( a \right) = f\left( { - a} \right)\\
Vậy\,f\left( a \right) = f\left( { - a} \right)\,khi: - 2 \le a \le 2\\
c)y = \sqrt {2 - x} + \sqrt {x + 2} \\
\Leftrightarrow {y^2} = 2 - x + 2\sqrt {2 - x} .\sqrt {x + 2} + x + 2\\
= 4 + 2\sqrt {4 - {x^2}} \\
{y^4} - 4\\
= \left( {{y^2} + 2} \right)\left( {{y^2} - 2} \right)\\
= \left( {4 + 2\sqrt {4 - {x^2}} + 2} \right)\left( {4 + 2\sqrt {4 - {x^2}} - 2} \right)\\
= \left( {2\sqrt {4 - {x^2}} + 6} \right)\left( {2\sqrt {4 - {x^2}} + 2} \right) > 0\\
Vậy\,{y^4} > 4
\end{array}$
