[tex]\text{Mai intai stabilim conditiile de existenta ale radicalului:}\\
2x-1\geq0 \Rightarrow x\geq \dfrac{1}{2}\\
2-x\geq 0\Rightarrow x\leq 2\\
C.E.:x\in \left[\dfrac{1}{2},2\right]\\
\sqrt{2x-1}+\sqrt{2-x}=2|()^2\\
2x-1+2-x+2\sqrt{(2x-1)(2-x)}=4\\
x+1+2\sqrt{(2x-1)(2-x)}=4\\
2\sqrt{(2x-1)(2-x)}=3-x|()^2\\
4(2x-1)(2-x)=9-6x+x^2\\
4(4x+x-2-2x^2)=9-6x+x^2\\
20x-8-8x^2=9-6x+x^2\\
9x^2-26x+17=0\\
9x^2-9x-17x+17=0\\
9x(x-1)-17(x-1)=0\\
(x-1)(9x-17)=0\Rightarrow x\in \left\{1,\dfrac{17}{9}\right\}[/tex]
