cho A = $\frac{√x +2}{√x-2}$ và B = ($\frac{√x}{x-4}$ +$\frac{1}{√x-2}$ ) : $\frac{√x+2}{x-4}$ vs x $\geq$ 0, x$\neq$ 4 a, rút gọn B b, tìm x nguyên

`#AkaShi`

a) Đk: `x >= 0` và `x \ne 4`

`((\sqrt{x})/(x-4)+1/(\sqrt{x}-2):(\sqrt{x}+2)/(x-4)`

`=((\sqrt{x})/(x-4)+(\sqrt{x}+2)/(x-4))*(x-4)/(\sqrt{x}+2)`

`=(\sqrt{x}+\sqrt{x}+2)/(x-4)*(x-4)/(\sqrt{x}+2)`

`=(2\sqrt{x}+2)/(x-4)*(x-4)/(\sqrt{x}+2)`

`=(2\sqrt{x}+2)/(\sqrt{x}+2)`

………….

b) `C=A*(B-2)`

`(\sqrt{x}+2)/(\sqrt{x}-2)*((2\sqrt{x}+2)/(\sqrt{x}+2)-2)`

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

`=(\sqrt{x}+2)/(\sqrt{x}-2)*((2\sqrt{x}+2)/(\sqrt{x}+2)-(2\sqrt{x}+4)/(sqrt{x}+2))`

`=(\sqrt{x}+2)/(\sqrt{x}-2)*(-2/(\sqrt{x}+2))`

`=-2/(\sqrt{x}-2)`

…………

Để `C ∈ Z` thì:

`2 \vdots \sqrt{x}-2`

`⇒\sqrt{x}-2 ∈Ư(2)={±1;±2}`

`* \sqrt{x}-2=1⇒x=9`

`* \sqrt{x}-2=-1⇒x=1`

`* \sqrt{x}-2=2⇒x=16`

`* \sqrt{x}-2=-2⇒x=0`

Vậy `x∈{9;1;16;0}` thì `C ∈ Z`