\(\left|x^2-3x+3\right|=3x-x^2-1\)
Do \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2-3x+3=3x-x^2-1\)
\(\Leftrightarrow2x^2-6x+4=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
Vậy \(x=1;2\)
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