\(x\left(3x-1\right)+6x-2=0\)
\(\Leftrightarrow3x^2-x+6x-2=0\)
\(\Leftrightarrow3x^2+5x-2=0\)
\(\Leftrightarrow3x^2+6x-x-2=0\)
\(\Leftrightarrow3x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
= 3x2-1+6x-2=0
= 3x2+6x-3=0
= 3x2+3x+3x-3=0
= (3x2+3x)+(3x-3)=0
= 3x(x+1)-3(x+1)=0
= (x+1)(3x-3)=0
= x+1=0 hoặc 3x-3=0
= x=1 hoặc 3x=3
x=1
vậy nghiệm của pt là S={1}
Ta có: \(x\left(3x-1\right)+6x-2=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-2\end{matrix}\right.\)
