Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{1+x}=b\end{matrix}\right.\) với \(a^2+b^2=2\)
\(M=\frac{\sqrt{1+ab}\left(b^3-a^3\right)}{2+ab}=\frac{\sqrt{1+ab}\left(b-a\right)\left(b^2+a^2+ab\right)}{2+ab}=\frac{\sqrt{\frac{a^2+b^2}{2}+ab}\left(b-a\right)\left(2+ab\right)}{2+ab}\)
\(=\left(b-a\right)\sqrt{\frac{a^2+b^2+2ab}{2}}=\frac{\left(b-a\right)\left(b+a\right)}{\sqrt{2}}\)
\(=\frac{b^2-a^2}{\sqrt{2}}=\frac{1+x-\left(1-x\right)}{\sqrt{2}}=x\sqrt{2}\)