Đáp án:
$x^2(x-4) +16-4x=0$
$⇔x^2(x-4) -4(x-4)=0$
$⇔(x-4)(x^2-4)=0$
$⇔(x-4)(x-2)(x+2)=0$
$⇔$\(\left[ \begin{array}{l}x-4=0\\x-2=0\\x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=4\\x=2\\x=-2\end{array} \right.\)
$\text{Vậy x ∈ { 4 ; 2 ; -2}}$
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$4xy -4x^2 -y^2+16$
$=16-(y^2-4xy+4x^2)$
$=16-(y-2x)^2$
$=(16-y+2x)(16+y-2x)$
