Đáp án:
Giải thích các bước giải:
$(x^3-x^2)^2-4x^2+8x-4=0$
$⇔x^4.(x-1)^2-4.(x^2-2x+1)=0$
$⇔x^4.(x-1)^2-4.(x-1)^2=0$
$⇔(x-1)^2.(x^4-4)=0$
$⇔(x-1)^2.(x^2-2).(x^2+2)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\x^2-2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x^2=2\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=±\sqrt{2}\end{array} \right.\)