$(3x-5)^{2}$-$(9x-1)^{2}$ =0 Tìm x

`(3x-5)^2 - (9x-1)^2=0`

`<=>[(3x-5)-(9x-1)][(3x-5)+(9x-1)]=0`

 `<=>(3x-5-9x+1)(3x-5+9x-1)=0`

`<=>(-6x-4)(12x-6)=0`

`<=>` $\left[\begin{matrix} -6x-4=0\\ 12x-6=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} -6x=4\\ 12x=6\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\frac{-6}{4}=\frac{-2}{3}\\ x=\frac{6}{12} = \frac{1}{2}\end{matrix}\right.$

Vậy `x=(-2)/3` hay `x=1/2`