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Đặt \(\sqrt{x}+\sqrt{1-x}=t>0\Rightarrow2\sqrt{x-x^2}=t^2-1\) (1)
Pt trở thành:
\(1+\frac{t^2-1}{3}=t\)
\(\Leftrightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
Thế vào (1): \(\left[{}\begin{matrix}2\sqrt{x-x^2}=0\\2\sqrt{x-x^2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-x^2=0\\4x^2-4x+9=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\left\{0;1\right\}\)