1. Solve the equation $2 x^{2}-3 x=2 x \sqrt{x^{2}-3 x}+1$ in $\mathbb{R}$.
Solution. $x^{2}+\left(x^{2}-3 x\right)=2 x \sqrt{x^{2}-3 x}+1$. If we set $x^{2}-3 x=y^{2}$, we get $x^{2}+y^{2}=2 x y+1, \quad$ i.e., $\quad(x-y)^{2}=1 . \quad$ From this, $\quad$ we obtain $\quad x-y= \pm 1, \quad$ i.e., $x-1=\sqrt{x^{2}-3 x}$ or $x+1=\sqrt{x^{2}-3 x}$. The first equation has no solution, while the solution to the second equation is $-\frac{1}{5}$.