Condition : \(0\le x\le1;x\ne\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\left(6x-3\right)\left(\sqrt{x}+\sqrt{1-x}\right)}{2x-1}=3+2\sqrt{x-x^2}\)

\(\Leftrightarrow3\cdot\left(\sqrt{x}+\sqrt{1-x}\right)=x+2\sqrt{x\left(1-x\right)}+\left(1-x\right)+2\)

\(\Leftrightarrow\left(\sqrt{x}+\sqrt{1-x}\right)^2-3\left(\sqrt{x}+\sqrt{1-x}\right)+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\sqrt{1-x}=1\\\sqrt{x}+\sqrt{1-x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x\left(1-x\right)}=0\\4x^2-4x+9=0\left(no-solution\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) So the expression has two solutions : x = 0; x = 1.