<=> [3(x-1)]2- [2(2x+1)]2= 0
<=> (3x-3)2 - (4x+2)2= 0
<=> (3x-3-4x-2)(3x-3+4x+2) = 0
<=> (-x-5)(7x-1) = 0
=> -x-5= 0 hoặc 7x-1= 0
=> x= -5 => x = 1/7
\(9\left(x-1\right)^2-4\left(2x+1\right)^2=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow9x^2-18x+9-16x^2-16x-4=0\)
\(\Leftrightarrow-7x^2-34x+5=0\)
\(\Leftrightarrow-7x^2+35x-x+5=0\)
\(\Leftrightarrow-7x\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(-7x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\-7x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1}{7}\end{matrix}\right.\)