\(M=\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2+4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)
Đặt \(\dfrac{4}{\sqrt{a}-2}=k\left(k\in Z\right)\)
\(\Rightarrow\sqrt{a}k-2k=4\\ \Rightarrow\sqrt{a}=\dfrac{4+2k}{k}\\ \Rightarrow\dfrac{2k+4}{k}\ge0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2k+4\ge0\\k>0\end{matrix}\right.\\\left\{{}\begin{matrix}2k+4\le0\\k>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}k\ge-2\\k>0\end{matrix}\right.\\\left\{{}\begin{matrix}k\le-2\\k< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k>0\\k< -2\end{matrix}\right.\)
\(\Rightarrow a=\left(\dfrac{2k+4}{k}\right)^2\)
Vậy để M nhận giá trị dương thì \(a=\left(\dfrac{2k+4}{k}\right)^2\) với \(k>0\text{ }hoặc\text{ }k\le-2\)
ĐK: a ≥0, a≠4
a) \(M=\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=\dfrac{\sqrt{a}-2+4}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)
Để M ∈ Z thì \(\dfrac{4}{\sqrt{a}-2}\) ∈ Z
⇒ 4 ⋮ \(\sqrt{a}-2\)
⇒ \(\sqrt{a}-2\) ∈ Ư(4)={-4; -2; -1;1; 2; 4}
Lập bảng
| \(\sqrt{a}-2\) | -4 | -2 | -1 | 1 | 2 | 4 |
| \(\sqrt{a}\) | -2 | 0 | 1 | 3 | 4 | 6 |
| a | 0 | 1 | 9 | 16 | 36 | |
| loại | tm | tm | tm | tm | tm |
Vậy khi a ∈ {0;1;9;16;36} thì M ∈ Z
