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$\#Lynn2007$
`-3x^2+2x+1=0`
`<=>-3x^2+3x-x+1=0`
`<=>-3x(x-1)-(x-1)=0`
`<=>(-3x-1)(x-1)=0`
`<=>-(3x+1)(x-1)=0`
`<=>(3x+1)(x-1)=0`
`<=>` \(\left[ \begin{array}{l}3x+1=0\\x-1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=1\end{array} \right.\)
Đáp án:
Giải thích các bước giải:
`- 3x^2 + 2x = 1 = 0`
` -3x^2 - x +3x +1 = 0 `
`( -3x^2 -x) +(3x+1)= 0`
` -x (3x+1) +(3x+1).1 = 0`
` (3x+1)(-x+1)= 0`
` (3x+1)(1-x) = 0`
`=>`\(\left[ \begin{array}{l}3x+1=0\\1-x= 0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}3x= -1\\x=1\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-\dfrac{1}{3}\\x=1\end{array} \right.\)
vậy ` x= -1/3; x=1`
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