\(\left(3x-2\right)\left(x^2+1\right)=3x-2\)
\(\left(3x-2\right)\left(x^2+1\right)-\left(3x-2\right)=0\)
\(\left(3x-2\right)\left(x^2+1-1\right)=0\)
\(x^2\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
(3x-2)(x2+1)=3x-2
(3x-2)(x2+1)-(3x-2)=0
(3x-2).((x2+1)-1)=0
3x-2=0 hoặc x2+0=0
x=2/3 hoặc x=0
\(\left(3x-2\right)\left(x^2+1\right)=3x-2\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+1\right)-\left(3x-2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)x^2=0\)
\(\Rightarrow\) \(x^2=0\) hoặc \(3x-2=0\)
x=0 \(3x=2\)
\(x=\dfrac{2}{3}\)
Vậy........