نکته: $\sqrt{{{x}^{2}}}=\left| x \right|=\left\{ \begin{align} & x \\ & -x \\ \end{align} \right.\begin{matrix} {} & {} \\ \end{matrix}\begin{matrix} x\ge 0 \\ x \lt 0 \\ \end{matrix}$.
نکته: ${{(a\pm b)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$
$A=\sqrt{2\sqrt{{{x}^{4}}+2{{x}^{2}}+1}-2\sqrt{{{x}^{4}}-2{{x}^{2}}+1}}\Rightarrow A=\sqrt{2\sqrt{{{({{x}^{2}}+1)}^{2}}}-2\sqrt{{{({{x}^{2}}-1)}^{2}}}}=\sqrt{2\left| {{x}^{2}}+1 \right|-2\left| {{x}^{2}}-1 \right|}$
$\xrightarrow{-1 \lt x \lt 0}A=\sqrt{2({{x}^{2}}+1)-2(1-{{x}^{2}})}=\sqrt{2{{x}^{2}}+2-2+2{{x}^{2}}}=\sqrt{4{{x}^{2}}}=\left| 2x \right|\xrightarrow{x \lt 0}-2x$
مهدی خلجی 
