Đáp án:
Bài 2 :
$a) x^2-5x=0$
$⇔x(x-5)=0$
$⇔$\(\left[ \begin{array}{l}x=0\\x-5=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=0\\x=5\end{array} \right.\)
Vậy x ∈ {0 ; 5}
$b) 5x.(x-2012)-x+2012=0$
$⇔5x(x-2012) -(x-2012)=0$
$⇔(x-2012)(5x-1)=0$
$⇔$\(\left[ \begin{array}{l}x-2012=0\\5x-1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2012\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy x ∈ {$2012 ; \dfrac{1}{5}$}
$c 2x.(x-5) -x.(2x+3)=26$
$⇔2x^2 -10x -2x^2-3x=26$
$⇔-13x=26$
$⇔x=26 : (-13)$
$⇔x=-2$
Vậy $x=-2$
