\(\left|x^2-3x+3\right|=\left|3x-x^2-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+3=3x-x^2-1\\x^2-3x+3=-3x+x^2+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+3-3x+x^2+1=0\\x^2-3x+3+3x-x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-6x+4=0\\0x+2=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
c2 : bình phương hai vế của phương trình , ta có :
( x2 - 3x + 3)2 = ( 3x - x2 - 1)2 ( sửa đề )
⇔ ( x2 - 3x + 3)2 - ( 3x - x2 - 1)2 = 0
⇔ ( x2 - 3x + 3 - 3x + x2 + 1)( x2 - 3x + 3 + 3x - x2 - 1) = 0
⇔ 2.( 2x2 - 6x + 4) = 0
⇔ 4( x2 - 3x + 2) = 0
⇔ x2 - x - 2x + 2 = 0
⇔ x( x - 1) - 2( x - 1) = 0
⇔ ( x - 1)( x - 2) = 0
⇔ x = 1 hoặc : x = 2
KL....