Ta có: \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(\Leftrightarrow x^2-16-6x+4=x^2-8x+16\)
\(\Leftrightarrow x^2-6x-12-x^2+8x-16=0\)
\(\Leftrightarrow2x-28=0\)
\(\Leftrightarrow2x=28\)
hay x=14
Vậy: S={14}
(x-4).(x+4)-2(3x-2)=(x-4)2
⇔x2 +4x-4x-16-6x+4=x2-8x+16
⇔x2-12-6x=x2-8x+16
⇔x2-x2-6x+8x=16+12
⇔2x=28
⇔x=14
\(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\\ \Leftrightarrow x^2-16-6x+4=x^2-4x+4\\\Leftrightarrow 2x=-16\\ \Leftrightarrow x=-8\\\)
