`ĐKXĐ:x\ge0,x\ne1`
`a,M={2}/{\sqrt{x}-1}+{2}/{\sqrt{x}+1}-{5-\sqrt{x}}/{x-1}`
`={2}/{\sqrt{x}-1}+{2}/{\sqrt{x}+1}-{5-\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={2(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}+{2(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}-{5-\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={2(\sqrt{x}+1)+2(\sqrt{x}-1)-(5-\sqrt{x})}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={2\sqrt{x}+2+2\sqrt{x}-2-5+\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={5\sqrt{x}-5}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={5(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`={5}/{\sqrt{x}+1}`
Vậy với `x\ge0,x\ne1` thì `M={5}/{\sqrt{x}+1}`
`b,x=4(TM)`
Thay `x=4` vào `M` có:
`M={5}/{\sqrt{4}+1}={5}/{2+1}={5}/{3}`
Vậy `x=4` thì `M=5/3`
`c,M\inZ⇔{5}/{\sqrt{x}+1}\inZ`
`⇔\sqrt{x}+1\inƯ(5)={-5;-1;1;5}`
`⇔\sqrt{x}\in{-6;-2;0;4}`
Mà `\sqrt{x}\ge0` với mọi `x\ge0`
`⇒\sqrt{x}\in{0;4}`
`⇔x\in{0;16}(TM)`
Vậy với `x\in{0;16}` thì `M\inZ`
