~Gửi bạn nhea~

Điều kiện: `x >= 0 ; x \ne 1`

`M=((\sqrtx+2)/(x+2\sqrtx+1)-(\sqrtx-2)/(x-1))*(\sqrtx+1)/(\sqrtx)`

`M=[(\sqrtx+2)/(\sqrtx+1)^2-(\sqrtx-2)/((\sqrtx+1)(\sqrtx-1))]*(\sqrtx+1)/(\sqrtx)`

`M=[((\sqrtx+2)(\sqrtx-1))/((\sqrtx+1)^2(\sqrtx-1))-((\sqrtx-2)(\sqrtx+1))/((\sqrtx+1)^2(\sqrtx-1))]*(\sqrtx+1)/(\sqrtx)`

`M=[((\sqrtx+2)(\sqrtx-1)-(\sqrtx-2)(\sqrtx+1))/((\sqrtx+1)^2(\sqrtx-1))]*(\sqrtx+1)/(\sqrtx)`

`M=(x-\sqrtx+2\sqrtx-2-(x+\sqrtx-2\sqrtx-2))/((\sqrtx+1)^2(\sqrtx-1))*(\sqrtx+1)/(\sqrtx)`

`M=(x-\sqrtx+2\sqrtx-2-x-\sqrtx+2\sqrtx+2)/((\sqrtx+1)^2(\sqrtx-1))*(\sqrtx+1)/(\sqrtx)`

`M=(2\sqrtx)/((\sqrtx+1)^2(\sqrtx-1))*(\sqrtx+1)/(\sqrtx)`

`M=2/((\sqrtx+1)(\sqrtx-1))`

`M=2/(x-1)`

.............................

`b)`

`2/(x-1)\ (x \ne 1)`

Để `M ∈ Z` thì:

`2 \vdots x-1`

`->x-1 ∈ Ư(2)={±1;±2}`

`**x-1=1→x=2\ (TM)`

`**x-1=-1→x=0\ (TM)`

`**x-1=2→x=3\ (TM)`

`->x-1=-2→x=-1\ (KTM)`

Vậy `x∈{2;3;0}` thì `M ∈ Z`

ĐKXĐ: `x>0`, `x\ne1`

`a,`

`M=({\sqrt{x}+2}/{x+2\sqrt{x}+1}-{\sqrt{x}-2}/{x-1}).{\sqrt{x}+1}/{\sqrt{x}}=[{\sqrt{x}+2}/{(\sqrt{x}+1)^2}-{\sqrt{x}-2}/{(\sqrt{x}+1)(\sqrt{x}-1)}].{\sqrt{x}+1}/{\sqrt{x}}=[{(\sqrt{x}+2)(\sqrt{x}-1)}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}-{(\sqrt{x}-2)(\sqrt{x}+1)}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}].{\sqrt{x}+1}/{\sqrt{x}}={(\sqrt{x}+2)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+1)}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}.{\sqrt{x}+1}/{\sqrt{x}}={x+2\sqrt{x}-\sqrt{x}-2-(x-2\sqrt{x}+\sqrt{x}-2)}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}.{\sqrt{x}+1}/{\sqrt{x}}={x+2\sqrt{x}-\sqrt{x}-2-x+2\sqrt{x}-\sqrt{x}+2}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}.{\sqrt{x}+1}/{\sqrt{x}}={2\sqrt{x}}/{(\sqrt{x}+1)^2(\sqrt{x}-1)}.{\sqrt{x}+1}/{\sqrt{x}}={2}/{(\sqrt{x}+1)(\sqrt{x}-1)}={2}/{x-1}`

Vậy với `x>0`, `x\ne1` thì `M={2}/{x-1}`

`b,`

`M∈Z⇔{2}/{x-1}∈Z⇔x-1∈Ư(2)={-2;-1;1;2}⇔x∈{-1;0;2;3}`

Kết hợp ĐKXĐ: `x>0`, `x\ne1`

`⇒x∈{2;3}`

Vậy với `x∈{2;3}` thì `M∈Z`