điều kiện \(x\in N;x\ne5\)
ta có : \(A\le0\Leftrightarrow\dfrac{x-2}{x-5}\le0\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x-5\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x-5\ge0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\ge5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2\le x\le5\\x\in\varnothing\end{matrix}\right.\) vậy \(2\le x\le5\) mà ta có \(x\in N\)
\(\Rightarrow x=2;x=3;x=4;x=5\) vậy \(x=2;x=3;x=4;x=5\)
Theo bài ra ta có :
\(A=\dfrac{x-2}{x-5}\left(x\in N;x\ne5\right)\\ A=\dfrac{x-5+3}{x-5}\\ A=1+\dfrac{3}{x-5}\)
Để \(A=1+\dfrac{3}{x-5}\le0\)
\(\text{thì : }\dfrac{3}{x-5}\le-1\\ \Leftrightarrow x-5\le-3\\ \Leftrightarrow x\le2\)
Vậy để \(A\le0\) thì \(x\le2\)
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