\(x^2-3x>0\Leftrightarrow x\left(x-3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)=> x>3 hoặc x < 0 thì mệnh dề đúng
b/ \(\sqrt{x}\ge x\)(đk:x ≥0)
\(\Leftrightarrow\sqrt{x}-x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)\ge0\)
=> 0 ≤ x ≤ 1
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