\(\Leftrightarrow\left(x-3-2x\right)\left(x-3+2x\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-1\\x=-3\end{matrix}\right.\)
=(x-3-4x)(x-3+4x)=0
=>TH1: x-4x-3=0
=>-3x=3
=>x=-1
TH2: x+4x-3=0
5x=3
x=\(\dfrac{3}{5}\)