Lời giải:
ĐKXĐ: \(x\geq \frac{1}{2}\)
Ta có: \(\sqrt{2x-1}+x^2-3x+1=0\)
\(\Leftrightarrow (x^2-2x+1)+(\sqrt{2x-1}-x)=0\)
\(\Leftrightarrow (x-1)^2+\frac{2x-1-x^2}{\sqrt{2x-1}+x}=0\)
\(\Leftrightarrow (x-1)^2-\frac{(x-1)^2}{\sqrt{2x-1}+x}=0\)
\(\Leftrightarrow (x-1)^2.\frac{\sqrt{2x-1}+x-1}{\sqrt{2x-1}+x}=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0(1)\\ \sqrt{2x-1}=1-x(2)\end{matrix}\right.\)
Với \((1)\Rightarrow x=1\) (thỏa mãn)
Với \((2)\Rightarrow \left\{\begin{matrix} 1-x\geq 0\\ 2x-1=(1-x)^2=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq 1\\ x^2-4x+2=0\end{matrix}\right.\Rightarrow x=2-\sqrt{2}\) (thỏa mãn)
Vậy.............