\(\left\{\begin{matrix}-1\le x\le1\\2-\sqrt{1-x^2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le1\) (*)
Đặt \(\left\{\begin{matrix}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a^2+b^2=2\\a,b\ge0\end{matrix}\right.\)
A tồn tại mọi x thuộc (*)
\(A=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{\frac{a^2-2ab+b^2}{2}}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)
\(A=\frac{\frac{\sqrt{2}}{2}\sqrt{\left(a-b\right)^2}\left(a+b\right)\left(2-ab\right)}{\left(2-ab\right)}\) Vơi đk (I)\(A=\frac{\sqrt{2}}{2}!a-b!\left(a+b\right)\)
\(\left[\begin{matrix}\left\{\begin{matrix}a\ge b\Leftrightarrow0\le x\le1\\A=\frac{\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{\sqrt{2}}{2}x\end{matrix}\right.\\\left\{\begin{matrix}a< b\Leftrightarrow-1\le x< 0\\A=\frac{-\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{-\sqrt{2}}{2}x\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow A=\frac{\sqrt{2}}{2}!x!\) Với x thỏa mãn đk (*)
