Giải thích các bước giải:
DKXD : $-1\le x\le 1$
$A=\dfrac{x}{\sqrt[]{1-x}+ \sqrt[]{1+x} }$
$\rightarrow 2A=\dfrac{2x}{\sqrt[]{1-x}+ \sqrt[]{1+x} }$
$\rightarrow 2A=\dfrac{1+x-(1-x)}{\sqrt[]{1-x}+ \sqrt[]{1+x} }$
$\rightarrow 2A=\dfrac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{\sqrt[]{1-x}+ \sqrt[]{1+x} }$
$\rightarrow 2A=\sqrt{1+x}-\sqrt{1-x}$
Vì $x\le 1\rightarrow 2A\le \sqrt{1+1}-\sqrt{1-1}=\sqrt{2}\rightarrow A\le \dfrac{\sqrt{2}}{2}$
$\rightarrow Max A=\dfrac{\sqrt{2}}{2}\rightarrow x=1$
Lại có $x\ge -1\rightarrow 2A\ge 0-\sqrt{1-(-1)}=-\sqrt{2}$
$\rightarrow MinA=-\sqrt{2}\rightarrow x=-1$
