\(3-x+\sqrt{x^2-3x+2}=0\)
\(\Leftrightarrow\sqrt{x^2-3x+2}=x-3\)
\(ĐKXĐ:x^2-3x+2\ge0\)
\(\Leftrightarrow\left(x-2\right).\left(x-1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\ge1\end{matrix}\right.\)
\(pt\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\x^2-3x+2=\left(x-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x^2-3x+2=x^2-6x+9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\x=\dfrac{7}{3}\left(tm\right)\end{matrix}\right.\)
Vậy x = \(\dfrac{7}{3}\)
\(3-x+\sqrt{x^2-3x+2}=0\)
\(pt\Leftrightarrow\dfrac{7}{3}-x+\sqrt{x^2-3x+2}-\dfrac{2}{3}=0\)
\(\Leftrightarrow\dfrac{7}{3}-x+\dfrac{x^2-3x+2-\dfrac{4}{9}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)
\(\Leftrightarrow-\left(x-\dfrac{7}{3}\right)+\dfrac{\left(x-\dfrac{7}{3}\right)\left(x-\dfrac{2}{3}\right)}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{3}\right)\left(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}\right)=0\)
Dễ thấy: \(-1+\dfrac{x-\dfrac{2}{3}}{\sqrt{x^2-3x+2}+\dfrac{2}{3}}>0\forall\left[{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\)
\(\Rightarrow x-\dfrac{7}{3}=0\Rightarrow x=\dfrac{7}{3}\)
