Đáp án:
$a) x^3 -0,25x =0$
$⇔ x(x^2-0,25) =0$
$⇔x(x-0,5)(x+0,5)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x-0,5=0\\x+0,5=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=0,5\\x=-0,5\end{array} \right.\)
$\text{Vậy x ∈ {0 ; -,5 ; -0,5} }$
$b) x^2 -10x = -25$
$⇔x^2-10x +25=0$
$⇔(x-5)^2=0$
$⇔x-5=0$
$⇔x=5$
Vậy $x=5$