Đáp án:
$\begin{array}{l}
Theo\,Cô - si:\\
Khi:a > 0;b > 0\\
\Rightarrow a + b \ge 2.\sqrt {a.b} \\
\Rightarrow \sqrt {ab} \le \dfrac{{a + b}}{2}\\
AD:Dkxd:0 < x < 2\sqrt 2 \\
x.\sqrt {8 - {x^2}} \le \dfrac{{{x^2} + 8 - {x^2}}}{2}\\
\Rightarrow x.\sqrt {8 - {x^2}} \le 4\\
\Rightarrow GTLN:x.\sqrt {8 - {x^2}} = 4\\
Khi:x = \sqrt {8 - {x^2}} \\
\Rightarrow {x^2} = 8 - {x^2}\\
\Rightarrow {x^2} = 4\\
\Rightarrow x = 2
\end{array}$