a.
\(\dfrac{3x\left(x-1\right)}{x^2+1}=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
3x ( x-1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy 2 pt trên tương đương