\(\sqrt{2x-1}+x^2-3x+1=0\) (ĐKXĐ: \(x\ge\dfrac{1}{2}\))

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

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

\(\Leftrightarrow2x-1=x^4+1+9x^2+2\left(x^2-3x-x^2.3x\right)\)

\(\Leftrightarrow2x-1=x^4+9x^2+1+2x^2-6x-6x^3\)

\(\Leftrightarrow x^4-6x^3+11x^2-8x+2=0\)

\(\Leftrightarrow x^4-x^3-5x^3+5x^2+6x^2-6x-2x+2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-5x^2+6x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-x^2-4x^2+4x+2x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^2-4x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[\left(x-2\right)^2-2\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-2\right)^2=2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=\sqrt{2}+2\left(TM\right)\\x=-\sqrt{2}+2\left(TM\right)\end{matrix}\right.\)

ĐKXĐ: `2x-1\ge0⇔x\ge1/2.`

`\sqrt{2x-1}+x^2-3x+1=0` 

`⇔\sqrt{2x-1}=-x^2+3x-1`

`⇔(\sqrt{2x-1})^2=(-x^2+3x-1)^2`

`⇔2x-1=x^4−6x^3+11x^2−6x+1` 

`⇔x^4-6x^3+11x^2-6x+1-2x+1=0`

`⇔x^4-6x^3+11x^2-8x+2=0`

`⇔x^4-2x^3+x^2-4x^3+8x^2 - 4x + 2x^2-4x+2=0`

`⇔ x^2(x^2-2x+1) -4x(x^2-2x+1)+2(x^2-2x+1)=0`

`⇔(x^2-2x+1)(x^2-4x+2)=0`

`⇔(x-1)^2(x^2-4x+4-2)=0`

`⇔(x-1)^2[(x-2)^2-(\sqrt{2})^2)=0`

`⇔(x-1)^2(x-2-\sqrt{2})(x-2+\sqrt{2})=0`

`⇒` \(\left[ \begin{array}{l}(x-1)^2=0\\x-2-\sqrt{2}=0\\x-2+\sqrt{2}=0\end{array} \right.\) 

`⇒` \(\left[ \begin{array}{l}x-1=0\\x=\sqrt{2}+2\\x=-\sqrt{2}+2\end{array} \right.\) 

`⇒` \(\left[ \begin{array}{l}x=1(tm)\\x=\sqrt{2}+2(tm)\\x=-\sqrt{2}+2(tm)\end{array} \right.\) 

Vậy tập nghiệm của phương trình: `S={1; ±\sqrt{2}+2}.`