\(\sqrt{2x-1}+x^2-3x+1=0\) (ĐKXĐ: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow\sqrt{2x-1}=-x^2+3x-1\)
\(\Leftrightarrow\left(\sqrt{2x-1}\right)^2=\left(-x^2+3x-1\right)^2=\left(x^2+1-3x\right)^2\)
\(\Leftrightarrow2x-1=x^4+1+9x^2+2\left(x^2-3x-x^2.3x\right)\)
\(\Leftrightarrow2x-1=x^4+9x^2+1+2x^2-6x-6x^3\)
\(\Leftrightarrow x^4-6x^3+11x^2-8x+2=0\)
\(\Leftrightarrow x^4-x^3-5x^3+5x^2+6x^2-6x-2x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-5x^2+6x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-x^2-4x^2+4x+2x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-4x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left[\left(x-2\right)^2-2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x-2\right)^2=2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=\sqrt{2}+2\left(TM\right)\\x=-\sqrt{2}+2\left(TM\right)\end{matrix}\right.\)
