\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\\ \Leftrightarrow\left(x+2\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
⇔(5x-4-49x)(5x-4+49x)=0
⇔(-44x-4)(54x-4)=0
⇒-44x-4=0 hoặc 54x-4=0
TH1:-44x-4=0 TH2:54x-4=0
⇔x=-1/11 ⇔x=2/27
Vậy xϵ{-1/11;2/27}
(5x-4-7x)(5x-4+7x)=0
(-2x-4)(12x-4)=0
⇒-2x-4=0 hoặc 12x-4=0
x=-2 x=1/3
