a/ \(3x\left(x-2\right)-x+2=0\)
\(\Rightarrow3x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
b/ \(4x\left(x-3\right)-2x+6=0\)
\(\Rightarrow4x\left(x-3\right)-\left(2x-6\right)=0\)
\(\Rightarrow4x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(4x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\4x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
c/ \(2x\left(x-4\right)+x-4=0\)
\(\Rightarrow\left(x-4\right)\left(2x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-4=0\\2x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d/ \(2x^3+4x=0\)
\(\Rightarrow x\left(2x^2+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\2x^2+4=0\Rightarrow x^2=-\dfrac{4}{2}=-2\end{matrix}\right.\)
Vì \(x^2=-2\) nên không xác định được x
Vậy x = 0
e/ \(3x^3-6x=0\)
\(\Rightarrow x\left(3x^2-6\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\3x^2-6=0\Rightarrow x^2=\dfrac{6}{3}=2\Rightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)