\(\left|x^2-3x+3\right|-x^2+3x-1\)
\(\text{Ta có : }x^2-3x+3\\ =x^2-3x+\dfrac{9}{4}+\dfrac{3}{4}\\ =\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{3}{4}\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
\(pt\Leftrightarrow x^2-3x+3-x^2+3x-1\\ \Leftrightarrow x^2-3x+3+x^2-3x+1=0\\ \Leftrightarrow2x^2-6x+4=0\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow x^2-2x-x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{1;2\right\}\)
