Question

choA= \(\sqrt{x}-x\)

tìm x để A>-2

Answer 1

Để A > -2 thì :

=> \(\sqrt{x}-x>-2\)

⇔ \(\sqrt{x}\left(1-\sqrt{x}\right)>-2\)

⇔\(\left\{{}\begin{matrix}\sqrt{x}>-2\\1-\sqrt{x}>-2\\\sqrt{x}< -2\\1-\sqrt{x}< -2\end{matrix}\right.\)  ⇔\(\left\{{}\begin{matrix}x>4\\x< 9\\x< 4\\x>9\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x>9\\x< 4\end{matrix}\right.\)

nếu mình sai thì sửa sau :>

Answer 2

Để A>-2 thì \(-x+\sqrt{x}+2>0\)

\(\Leftrightarrow x-\sqrt{x}-2>0\)

=>x>4