\(M=\dfrac{7\sqrt{a}-2}{2\sqrt{a}+1}\left(đk:a\ge0\right)=\dfrac{3\left(2\sqrt[]{a}+1\right)+\sqrt{a}-5}{2\sqrt{a}+1}=3+\dfrac{\sqrt{a}-5}{2\sqrt{a}+1}\)
Để \(M\in Z,M>0\) thì \(\sqrt{a}-5\ge0\Leftrightarrow a\ge25\) và:
\(\left\{{}\begin{matrix}\sqrt{a}-5⋮2\sqrt{a}+1\\2\sqrt{a}+1⋮2\sqrt{a}+1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2\sqrt{a}-10⋮2\sqrt{a}+1\\2\sqrt{a}+1⋮2\sqrt{a}+1\end{matrix}\right.\)
\(\Rightarrow\left(2\sqrt{a}+1\right)-\left(2\sqrt{a}-10\right)⋮2\sqrt{a}+1\)
\(\Rightarrow11⋮2\sqrt{a}+1\Rightarrow2\sqrt{a}+1\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)
Do \(\sqrt{a}\ge0\forall a\)
\(\Rightarrow\sqrt{a}\in\left\{0;5\right\}\)
\(\Rightarrow a\in\left\{0\left(loại\right);25\left(nhận\right)\right\}\)
