Đặt \(a=\sqrt{2+x};\text{ }b=\sqrt{2-x}\Rightarrow a^2+b^2=4\)
\(A=\frac{\sqrt{2+ab}\left(a^3-b^3\right)}{a^2+b^2+ab}=\frac{\sqrt{2+ab}\left(a-b\right)\left(a^2+b^2+ab\right)}{a^2+b^2+ab}=\left(a-b\right)\sqrt{\frac{a^2+b^2}{2}+ab}\)
\(=\left(a-b\right)\sqrt{\frac{\left(a+b\right)^2}{2}}=\frac{\left(a-b\right)\left(a+b\right)}{\sqrt{2}}\)
\(=\frac{a^2-b^2}{\sqrt{2}}=\frac{\left(2+x\right)-\left(2-x\right)}{\sqrt{2}}=\frac{2x}{\sqrt{2}}=x\sqrt{2}\)