*Bài làm:
~Ta có: \(A\) = \(\frac{x^2+1}{2x^2+6x}\) = \(\frac{x^2+1}{x^2+1+x^2+6x-1}\)
= \(\frac{x^2+1}{x^2+1}\) + \(\frac{1}{x^2+6x-1}\) = 1 + \(\frac{1}{x^2+6x-1}\)
⇒Để \(A\) âm thì: 1 + \(\frac{1}{x^2+6x-1}\) < 0
⇒ \(\frac{1}{x^2+6x-1}\) < -1
⇒ 1 < -1.(\(x^2+6x-1\))
⇒ 1 < \(-\left(x^2\right)-6x+1\)
⇒ 0 < \(-\left(x^2\right)-6x\)
⇒ 0 < \(x\left(-x-6\right)\)
~Mà \(x\) \(\in\) \(Z\) ( Theo đề bài ).
⇒ \(x\) ; \(-x-6\) cùng dấu.
⇒ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\-x-6>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\-x-6< 0\end{matrix}\right.\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\-x>6\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\-x< 6\end{matrix}\right.\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 6\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x>6\end{matrix}\right.\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}0< x< 6\left(Chọn\right)\\6< x< 0\left(Loại\right)\end{matrix}\right.\)
⇒ \(x\) \(\in\) \(\left\{1;2;3;4;5\right\}\) (do \(x\) \(\in\) \(Z\) ).
*Vậy: \(x\) \(\in\) \(\left\{1;2;3;4;5\right\}\) thỏa mãn đề.
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