a) Ta có: \(2x^2=x\)
\(\Leftrightarrow2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;\frac{1}{2}\right\}\)
b) Ta có: \(x^3+9x=6x^2\)
\(\Leftrightarrow x^3+9x-6x^2=0\)
\(\Leftrightarrow x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy: S={0;3}
a) \(2x^2=x\)
\(\Rightarrow2x^2-x=0\)
\(\Rightarrow x\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy:............
b) \(x^3+9x=6x^2\)
\(\Rightarrow x^3+9x-6x^2=0\)
\(\Rightarrow x^3-6x^2+9x=0\)
\(\Rightarrow x\left(x^2-6x+9\right)=0\)
\(\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy:...................
