\(x\left(2-x\right)+\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow2x-x^2+x^2-16=0\)
\(\Leftrightarrow2x=16\)
\(\Leftrightarrow x=8\)
Vậy: \(S=\left\{8\right\}\)
x ( 2 - x ) + ( x2 - 16 ) = 0
-x2 + 2x + x2 - 16 = 0
2x - 16 = 0
=> 2 ( x - 8 ) = 0
=> x - 8 = 0
x = 8
Vậy x = 8