pt đã cho
\(\Leftrightarrow2x^2-5x+2-\left(x-2\right)\sqrt{x^2-x+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-\sqrt{x^2-x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\2x-1-\sqrt{x^2-x+1}=0\end{matrix}\right.\)
(*) \(2x-1-\sqrt{x^2-x+1}=0\) (đk: \(x\ge\dfrac{2+\sqrt{3}}{4}\))
Ta thấy \(2x-1+\sqrt{x^2-x+1}\ne0\) với mọi \(x\ge\dfrac{2+\sqrt{3}}{4}\), (*) tương đương:
\(\dfrac{\left(2x-1\right)^2-\left(x^2-x+1\right)}{2x-1+\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\dfrac{3x\left(x-1\right)}{2x-1+\sqrt{x^2-x+1}}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\\dfrac{3}{2x-1+\sqrt{x^2-x+1}}=0\left(vôlí\right)\end{matrix}\right.\)
Vậy pt đã cho có tập nghiệm \(S=\left\{1;2\right\}\)
