1: \(\left|x^2-3x+2\right|=x+2\)
=>\(\begin{cases}x+2\ge0\\ \left(x^2-3x+2\right)^2=\left(x+2\right)^2\end{cases}\)
=>\(\begin{cases}x\ge-2\\ \left(x^2-3x+2-x-2\right)\left(x^2-3x+2+x+2\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-2\\ \left(x^2-4x\right)\left(x^2-2x+4\right)=0\end{cases}\Rightarrow\begin{cases}x\ge-2\\ x\left(x-4\right)=0\end{cases}\)
=>x∈{0;4}
2: \(\left|3x^2-4x+1\right|-\left|3x+1\right|=0\)
=>\(\left|3x^2-4x+1\right|=\left|3x+1\right|\)
=>\(\left[\begin{array}{l}3x^2-4x+1=3x+1\\ 3x^2-4x+1=-3x-1\end{array}\right.\Rightarrow\left[\begin{array}{l}3x^2-7x=0\\ 3x^2-x+2=0\left(loại\right)\end{array}\right.\)
=>\(3x^2-7x=0\)
=>x(3x-7)=0
=>x=0 hoặc x=7/3
