Đáp án:
Phía dưới
Giải thích các bước giải:
` a) 4x+12x^2=0`
`⇔4x(1+3x)=0`
`⇔`\(\left[ \begin{array}{l}4x=0\\1+3x=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy `x=0` hoặc `x=-1/3`
`b) 3(x+1)^2-x-1=0`
`⇔3(x+1)^2-(x+1)=0`
`⇔[3(x+1)-1](x+1)=0`
`⇔(3x+3-1)(x+1)=0`
`⇔(3x+2)(x+1)=0`
`⇔`\(\left[ \begin{array}{l}3x+2=0\\x+1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=-1\end{array} \right.\)
Vậy `x=-2/3` hoặc` x=-1`
`c) 2x(x-2)^2-x(x-2)=0`
`⇔x(x-2)[2(x-2)-1]=0`
`⇔x(x-2)(2x-4-1)=0`
`⇔x(x-2)(2x-5)=0`
`⇔`\(\left[ \begin{array}{l}x=0\\x-2=0\\2x-5=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=0\\x=2 \\x=\dfrac{5}{2}\end{array} \right.\)
Vậy `x=0` hoặc `x=2` hoặc `x=5/2`
