a)\(2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=0\) ; \(x=\dfrac{3}{2}\)
b)hình như đề sai
c) \(x^2=\left(2x-1\right)^2\)
\(\Leftrightarrow x^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(x-2x+1\right)\left(x+2x-1\right)=0\)
\(\Leftrightarrow\left(-x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x+1=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-x=-1\\3x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x=1\) ; \(x=\dfrac{1}{3}\)
d) \(4x^3-x=0\)
\(\Leftrightarrow x\left(4x^2-1\right)=0\)
\(\Leftrightarrow x\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy \(x=0\) ; \(x=\dfrac{1}{2}\) ; \(x=\dfrac{-1}{2}\)