Question

Tìm x biết:

a) \(\frac{x+5}{x-4}>0\)

b) \(\frac{x+8}{x^2+7}< 0\)

Giúp mình gấp với nhé!

Answer 1

a) \(\frac{x+5}{x-4}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+5>0\\x-4>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+5< 0\\x-4< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-5\\x>4\end{matrix}\right.\\\left\{{}\begin{matrix}x< -5\\x< 4\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -5\end{matrix}\right.\)

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b) \(\frac{x+8}{x^2+7}< 0\)

Ta có \(x^2+7>0\Leftrightarrow x+8< 0\Leftrightarrow x< -8\)

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