Question

[nguyễn thị thu hường]

Answer 1

[2611]

`a)` Với `x > 0,x \ne 1` có:

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

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

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

`A=[2\sqrt{x}]/[\sqrt{x}(x-1)]`

`A=2/[x-1]`

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`2)A in ZZ<=>2/[x-1] in ZZ`

       `=>x-1 in Ư_2`

 Mà `Ư_2={+-1;+-2}`

 `@x-1=1=>x=2` (t/m)

 `@x-1=-1=>x=0` (ko t/m)

 `@x-1=2=>x=3` (t/m)

  `@x-1=-2=>x=-1` (ko t/m)

Vậy `x in {2;3}`