\(3x^2+3x-6=0\)
\(\Leftrightarrow3x^2+6x-3x+6=0\)
\(\Leftrightarrow\left(3x^2-3x\right)+\left(6x+6\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+6\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+6=0\Rightarrow3x=6\Rightarrow x=2\\x-1=0\Rightarrow x=1\end{cases}}\)
Vậy x=2 và x=1
cau tra loi chuan ne ban
\(3x^2+3x-6=0\)
\(3x^2+6x-3x-6=0\)
\(3x\left(x-1\right)+6\left(x-1\right)=0\)
\(\left(3x+6\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+6=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)
vay x=-2 va x=1
\(3x^2-3x-6=0\)
\(=3x^2+6x-3x-6=0\)
\(=3x\left(x-1\right)+6\left(x-1\right)\)
\(=\left(3x+6\right)\left(x-1\right)=0\)
\(\Rightarrow\)\(\orbr{\begin{cases}3x+6=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6:3=2\\x=1\end{cases}}}\)
Vậy \(x=\left(2;1\right)\)
