\(M=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\)
Để \(M\in Z\Leftrightarrow7⋮\sqrt{x}-2\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-2=1\\\sqrt{x}-2=-1\\\sqrt{x}-2=7\\\sqrt{x}-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=9\\\sqrt{x}=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\\x=81\\loại\end{matrix}\right.\)
\(M=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{7}{\sqrt{x}-2}\)
\(=1+\dfrac{7}{\sqrt{x}-2}\)
Để \(M\in Z\Leftrightarrow\dfrac{7}{\sqrt{x}-2}\Leftrightarrow\left(\sqrt{x}-2\right)\inƯ\left(7\right)\)
\(\Leftrightarrow\sqrt{x}-2=\left\{\pm1;\pm7\right\}\) \(\Leftrightarrow\sqrt{x}=\left\{-5;1;3;9\right\}\)
\(\Leftrightarrow x=\left\{1;9;81\right\}\)
Vậy.....................
\(M=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}+2}\)
Để M nhận giá trị nguyên <=> \(\sqrt{x}+2\inƯ_{\left(3\right)}=\left\{\pm1;\pm3\right\}\)
\(\rightarrow\sqrt{x}=\left\{-1;-3;1;-5\right\}\)
\(\Rightarrow x=1\)
Vậy có 1 giá trị của x = 1 để M nhận giá trị nguyên.
