Câu `2`
`a)3x^2+4x=0`
`⇔x(3x+4)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\3x+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{4}{3}\end{array} \right.\)
Vậy `S={0;-4/3}`
`b)(2x+1)^2=(x-1)^2`
`⇔(2x+1)^2-(x-1)^2=0`
`⇔(2x+1-x+1)(2x+1+x-1)=0`
`⇔3x(x+2)=0`
`⇔` \(\left[ \begin{array}{l}3x=0\\x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\)
Vậy `S={0;-2}`
`c)x^2-x-6=0`
`⇔x^2-3x+2x-6=0`
`⇔x(x-3)+2(x-3)=0`
`⇔(x-3)(x+2)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\x+2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\\x=-2\end{array} \right.\)
Vậy `S={3;-2}`
