Đáp án:
Giải thích các bước giải:
a) `6x(x-2012)+x-2012=0`
`⇔ (6x+1)(x-2012)=0`
`⇔` \(\left[ \begin{array}{l}6x+1=0\\x-2012=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=2012\end{array} \right.\)
Vậy `S={-\frac{1}{6};2012}`
b) `x^2+8x=0`
`⇔ x(x+8)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x+8=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.\)
Vậy `S={0;-8}`
c) `x^2-7x=0`
`⇔ x(x-7)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x-7=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=7\end{array} \right.\)
Vậy `S={0;7}`
