$\text{1)}\\M=\dfrac{2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+1}-\dfrac{5-\sqrt{x}}{x-1}\\\hspace{0,5cm}=\dfrac{2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+1}-\dfrac{5-\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\\\hspace{0,5cm}=\dfrac{2(\sqrt{x}+1)+2(\sqrt{x}-1)-5+\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\\\hspace{0,5cm}=\dfrac{2\sqrt{x}+2+2\sqrt{x}-2-5+\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}\\\hspace{0,5cm}=\dfrac{5\sqrt{x}-5}{(\sqrt{x}-1)(\sqrt{x}+1)}\\\hspace{0,5cm}=\dfrac{5(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\\\hspace{0,5cm}=\dfrac{5}{\sqrt{x}+1}\\\text{2)}\\M=\dfrac{5}{\sqrt{x}+1}\,\,\,(x\ge0,\,x\ne1)\\\text{Với }x=4\to M=\dfrac{5}{\sqrt{4}+1}=\dfrac{5}{3}\\\text{3)}\\\text{Để M nguyên }\to 5\vdots(\sqrt{x}+1)\\\to \sqrt{x}+1\in{Ư(5)=\{\pm1;\,\pm5\}}\\\to \left[\begin{array}{}\hspace{-0,3cm}\sqrt{x}+1=1\\\sqrt{x}+1=-1\\\hspace{-0,4cm}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{array}\right.\\\to \left[\begin{array}{}\hspace{-0,3cm}\sqrt{x}=0 \to x=0\text{ (thoả mãn)}\\\hspace{-3cm}\sqrt{x}=-2\text{ (loại)}\\\hspace{0cm}\sqrt{x}=4 \to x=16\text{ (thoả mãn)}\\\hspace{-3cm}\sqrt{x}=-6 \text{ (loại)}\end{array}\right.\\\text{Vậy }x=\{0,\,16\}$
