\(3x^2+2x-1=0\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)
Vậy...
Ta có :
\(3x^2+2x-1=0\)
\(\Leftrightarrow\)\(3x^2+3x-x-1=0\)
\(\Leftrightarrow\)\(\left(3x^2+3x\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\)\(3x.\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\)\(\left(3x-1\right).\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=1\\x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}}\)
Vậy \(x=-1\) hoặc \(x=\frac{1}{3}\)
