Đáp án+Giải thích các bước giải:
a,
$x(x+1)-x(x-3)=8$
$⇔x^2+x-x^2+3x=8$
$⇔4x=8$
$⇔x=2$
Vậy $x=2$
b,
$x^2-6x+8=0$
$⇔x^2-4x-2x+8=0$
$⇔x(x-4)-2(x-4)=0$
$⇔(x-2)(x-4)=0$
$⇔\left[\begin{matrix}x-2=0\\x-4=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=2\\x=4\end{matrix}\right.$
Vậy `S={2;4}`
c,
$2x^2+2x+\dfrac{1}{2}=0$
$⇔2(x^2+x+\dfrac{1}{4})=0$
$⇔2(x+\dfrac{1}{2})^2=0$
$⇔x+\dfrac{1}{2}=0$
$⇔x=\dfrac{-1}{2}$
Vậy `S={\frac{-1}{2}}`