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