Đáp án:
$a, (x-1)^2 -x(x+2)=-3+4x$
$⇔ x^2-2x+1 -x^2-2x =-3+4x$
$⇔x^2-x^2-2x-2x-4x=-3-1$
$⇔-8x=-4$
$⇔x=-4 : (-8)$
$⇔x=\dfrac{1}{2}$
Vậy $x=\dfrac{1}{2}$
$b) (x-2).(x^2+2x+4) = 19$
$⇔ x^3 +2x^2+4x-2x^2-4x-8=19$
$⇔x^3 =19 +8$
$⇔x^3=27$
$⇔x=3$
Vậy $x=3$
$c) (x-1)^2 -(2x+3)^2=0$
$⇔ (x-1-2x-3)(x-1+2x+3)=0$
$⇔(-x-4).(3x+2)=0$
$⇔$\(\left[ \begin{array}{l}-x-4=0\\3x+2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-4\\x=-\dfrac{2}{3}\end{array} \right.\)
$\text{Vậy x ∈ {$-4; -\dfrac{2}{3}$}}$
$d) (x-1)^3 -x(x-2)^2-(x-2)=0$
$⇔ x^3 -3x^2+3x-1 -x^3+4x^2-4x -x+2=0$
$⇔x^2-2x+1=0$
$⇔(x-1)^2=0$
$⇔x-1=0$
$⇔x=1$
Vậy $x=1$
