\(3x^2+2x-1=0\)
\(\Rightarrow3x^2+3x-x-1=0\)
\(\Rightarrow3x.\left(x+1\right)-\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right).\left(3x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-1;\dfrac{1}{3}\right\}\)
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Bài làm:
\(3x^2+2x-1=0\\ < =>x^2+2x^2+2x+1-2=0\\ < =>\left(x^2+2x+1\right)+\left(2x^2-2\right)=0\\ < =>\left(x+1\right)^2+2\left(x-1\right)\left(x+1\right)=0\\ < =>\left(x+1\right)\left(x+1+2\left(x-1\right)\right)=0\\ < =>\left(x+1\right)\left(x+1+2x-2\right)=0\\ < =>\left(x+1\right)\left(3x-1\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
