Question

Giải BĐT sau:

\(x^2>2x\)

Answer 1

\(x^2>2x\\ \Leftrightarrow x^2-2x>0\)
\(\Leftrightarrow x\left(x-2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 0\end{matrix}\right.\)

Answer 2

Có \(x^2>2x\Leftrightarrow x^2-2x>0\Leftrightarrow x\left(x-2\right)>0\)

\(\Leftrightarrow[\begin{matrix}\left\{{}\begin{matrix}x>0\\x-2>0\Leftrightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-2< 0\Leftrightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)