Làm dùm em với mn :<

$\begin{array}{l}M = \dfrac{a + 1}{\sqrt a} + \dfrac{a\sqrt a -1}{a- \sqrt a} + \dfrac{a^2 – a\sqrt a + \sqrt a – 1}{\sqrt a – a\sqrt a}\\ a)\,\,\text{M có nghĩa}\,\Leftrightarrow \begin{cases}a \geq 0\\\sqrt a \ne 0\\a – \sqrt a \ne 0\\\sqrt a – a\sqrt a \ne 0\end{cases}\Leftrightarrow \begin{cases}a > 0\\a \ne 1\end{cases}\\ b)\\M = \dfrac{a + 1}{\sqrt a} + \dfrac{(\sqrt a – 1)(a + \sqrt a + 1)}{\sqrt a(\sqrt a-1)} + \dfrac{a^2 -1 – \sqrt a(a – 1)}{\sqrt a(1 – a)}\\ \to M = \dfrac{a + 1}{\sqrt a} + \dfrac{a + \sqrt a + 1}{\sqrt a} – \dfrac{a + 1 -\sqrt a}{\sqrt a}\\ \to M = \dfrac{a + 1 + a + \sqrt a + 1 – a + \sqrt a – 1}{\sqrt a}\\ \to M = \dfrac{a + 2\sqrt a}{\sqrt a}\\ \to M = \sqrt a + 2\\ c)\,\,N = \dfrac{6}{M} = \dfrac{6}{\sqrt a + 2}\\ N \in \Bbb Z \Leftrightarrow \sqrt a + 2 \in Ư(6) = \left\{-6;-3;-2;-1;1;2;3;6\right\}\\ \text{Ta lại có:}\\ \sqrt a > 0\\ \to \sqrt a + 2 > 2\\ \to \sqrt a + 2 = \left\{3;6\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r|}
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\sqrt a + 2 &3&6\\
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\quad \sqrt a&1&4\\
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\quad \,\,a&1&16\\
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\end{array}\\
Vậy\,\,a = \left\{1;16\right\}\end{array}$