Tập nghiệm của \(\left|x^{2...

A

- Hướng dẫn giải

Ta có

\(\left|x^{2}-3 x+3\right|=-x^{2}+3 x-1 \Leftrightarrow\left\{\begin{array}{l} -x^{2}+3 x-1 \geq 0 \\ {\left[\begin{array}{l} x^{2}-3 x+3=-x^{2}+3 x-1 \\ x^{2}-3 x+3=x^{2}-3 x+1 \end{array}\right.} \end{array}\right.\)

\(\Leftrightarrow\left\{\begin{array}{l} -x^{2}+3 x-1 \geq 0 \\ {\left[\begin{array}{l} 2 x^{2}-6 x+4=0 \\ 3=1 \quad(L) \end{array}\right.} \end{array} \Leftrightarrow\left\{\begin{array}{l} -x^{2}+3 x-1 \geq 0 \\ (x-2)(x-1)=0 \end{array}\right.\right.\)

\(\Leftrightarrow\left\{\begin{array}{l} -x^{2}+3 x-1 \geq 0\left(^{*}\right) \\ {\left[\begin{array}{l} x=2 \\ x=1 \end{array}\right.} \end{array} \Leftrightarrow\left[\begin{array}{l} x=1 \\ x=2 \end{array}\left(t/ m\left(^{*}\right)\right)\right.\right.\)

Vậy \(\mathrm{S}=\{1 ; 2\}\)