ĐKXĐ: \(0\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\) ta được hệ:
\(\left\{{}\begin{matrix}a^2+b^2=1\\\frac{2a^3}{a+b}+ab=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a^3+ab\left(a+b\right)=a+b\\a^2+b^2=1\end{matrix}\right.\)
\(\Rightarrow2a^3+a^2b+ab^2=\left(a+b\right)\left(a^2+b^2\right)\)
\(\Rightarrow2a^3+a^2b+ab^2=a^3+ab^2+ab^2+b^3\)
\(\Leftrightarrow a^3=b^3\)
\(\Leftrightarrow a=b\Leftrightarrow\sqrt{x}=\sqrt{1-x}\)
\(\Leftrightarrow x=1-x\Leftrightarrow x=\frac{1}{2}\)