1) Ta có: |x-2|+|x+2|=|2x|
\(\Leftrightarrow x-2+x+2+2\left|x^2-4\right|=2x\)
\(\Leftrightarrow2x+2\left|x^2-4\right|=2x\)
\(\Leftrightarrow2\left|x^2-4\right|=0\)
\(\Leftrightarrow\left|x^2-4\right|=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: S={2;-2}
2) Ta có: |x-2|+|3x-9|=|x+1|
\(\Leftrightarrow x-2+3x-9+2\left|\left(x-2\right)\left(3x-9\right)\right|=x+1\)
\(\Leftrightarrow4x-11+2\left|\left(x-2\right)\left(3x-9\right)\right|=x+1\)
\(\Leftrightarrow4x-11+2\left|\left(x-2\right)\left(3x-9\right)\right|-x-1=0\)
\(\Leftrightarrow2\left|\left(x-2\right)\left(3x-9\right)\right|+3x-12=0\)
\(\Leftrightarrow2\left|\left(x-2\right)\left(3x-9\right)\right|=12-3x\)
\(\Leftrightarrow4\left(x-2\right)\left(3x-9\right)=\left(12-3x\right)^2\)
\(\Leftrightarrow4\left(3x^2-9x-6x+18\right)=144-72x+9x^2\)
\(\Leftrightarrow4\left(3x^2-15x+18\right)=9x^2-72x+144\)
\(\Leftrightarrow12x^2-60x+72-9x^2+72x-144=0\)
\(\Leftrightarrow3x^2+12x-72=0\)
\(\Leftrightarrow3\left(x^2+4x-24\right)=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot2+4-28=0\)
\(\Leftrightarrow\left(x+2\right)^2=28\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2\sqrt{7}\\x+2=-2\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{7}-2\\x=-2\sqrt{7}-2\end{matrix}\right.\)
Vậy: \(S=\left\{2\sqrt{7}-2;-2\sqrt{7}-2\right\}\)
