\(\sqrt{2x-1}+x^2-3x+1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\sqrt{2x-1}-x=0\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{2x-1-x^2}{\sqrt{2x-1}+x}=0\)
\(\Leftrightarrow\left(x-1\right)^2-\dfrac{\left(x-1\right)^2}{\sqrt{2x-1}+x}=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(1-\dfrac{1}{\sqrt{2x-1}+x}\right)=0\)
Trường hợp 1:
\(\left(x-1\right)^2=0\)
\(\Leftrightarrow x=1\)
Trường hợp 2:
\(1-\dfrac{1}{\sqrt{2x-1}+x}=0\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2x-1}+x}=\dfrac{1}{1}\)
\(\Leftrightarrow\sqrt{2x-1}=1-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\2x-1=1-2x+x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1\ge x\\x^2-4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left[{}\begin{matrix}x=2+\sqrt{2}\left(l\right)\\x=2-\sqrt{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy phương trình có 2 nghiệm phân biệt . . .
Ta tách:
\(\Leftrightarrow\sqrt{2x-1}+x^2-x-\left(2x-1\right)=0\)
\(\Leftrightarrow x^2-x=\left(2x-1\right)-\sqrt{2x-1}\)
Đặt \(t=\sqrt{2x-1}\) ta có:
\(x^2-x=t^2-t\)
\(\Leftrightarrow\left(x-t\right)\left(x+t-1\right)=0\)
Ta đưa về 2 phương trình:
\(x=\sqrt{2x-1}\) hoặc \(x+\sqrt{2x-1}=1\)
