Lời giải.

`\text{ĐKXĐ}:x>0,x\ne4`

`a)P=(1/{\sqrt{x}+2}+1/{\sqrt{x}-2}).{\sqrt{x}-2}/\sqrt{x}`

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

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

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

`P=2/{\sqrt{x}+2}.`

Vậy với `x>0,x\ne4` thì `P=2/{\sqrt{x}+2}.`

`b)` Ta có: `7P=7. 2/{\sqrt{x}+2}= 14/ {\sqrt{x}+2}.`

`{7P}/3=7P. 1/3= 14/ {\sqrt{x}+2} . 1/3 = 14/{3(\sqrt{x}+2)}=14/{3\sqrt{x}+6}`

Với `x>0=>3\sqrt{x}>0=>3\sqrt{x}+6>6`

`=>14/{3\sqrt{x}+6}<14/6≈2,33` `(1)`

Mà `3\sqrt{x}+6>6>0,14>0=>14/{3\sqrt{x}+6}>0` `(2)`

Từ `(1)` và `(2)=>0<{7P}/3<2,33`

Mà `{7P}/3∈ZZ=>{7P}/3∈{1;2}`

`+)` Xét `{7P}/3=1<=>14/{3\sqrt{x}+6}=1`

`<=>14=3\sqrt{x}+6`

`<=>8=3\sqrt{x}`

`<=>8/3=\sqrt{x}`

`<=>x=(8/3)^2=64/9(tmdk)`

`+)` Xét `{7P}/3=2<=>14/{3\sqrt{x}+6}=2`

`<=>14=6\sqrt{x}+12`

`<=>2=6\sqrt{x}`

`<=>1/3=\sqrt{x}`

`<=>x=(1/3)^2=1/9(tmdk)`

Vậy `x∈{1/9;64/9}` thì `{7P}/3∈ZZ.`

a/ ĐKXĐ: $x>0,x\ne 4$

$P=\big(\dfrac{1}{\sqrt x+2}+\dfrac{1}{\sqrt x-2}\big).\dfrac{\sqrt x-2}{\sqrt x}\\=\big(\dfrac{\sqrt x-2}{(\sqrt x+2)(\sqrt x-2)}+\dfrac{\sqrt x+2}{(\sqrt x+2)(\sqrt x-2)}\big).\dfrac{\sqrt x-2}{\sqrt x}\\=\dfrac{\sqrt x-2+\sqrt x+2}{(\sqrt x+2)(\sqrt x-2)}.\dfrac{\sqrt x-2}{\sqrt x}\\=\dfrac{2\sqrt x.(\sqrt x-2)}{(\sqrt x+2)(\sqrt x-2)\sqrt x}\\=\dfrac{2}{\sqrt x+2}$

Vậy $P=\dfrac{2}{\sqrt x+2}$ với $x>0,x\ne 4$

b/ $\dfrac{7P}{3}\\=\dfrac{7.\dfrac{2}{\sqrt x+2}}{3}\\=\dfrac{14}{\sqrt x+2}.\dfrac{1}{3}\\=\dfrac{14}{3(\sqrt x+2)}$

Để $\dfrac{7P}{3}∈\Bbb Z\\→\dfrac{14}{3(\sqrt x+2)}∈\Bbb Z\\→14\vdots 3(\sqrt x+2)\\→3(\sqrt x+2)∈Ư(14)=\{±1;±2;±7;±14\}$

Vì $\sqrt x≥0$

$→\sqrt x+2\ge 2\\→3(\sqrt x+2)\ge 6>0\\→3(\sqrt x+2)∈\{7;14\}$

Ta có bảng:

$\begin{array}{|c|c|c|}\hline 3(\sqrt x+2)&7&14\\\hline \sqrt x+2&\dfrac{7}{3}&\dfrac{14}{3}\\\hline \sqrt x&\dfrac{1}{3}&\dfrac{8}{3}\\\hline x&\dfrac{1}{9}&\dfrac{64}{9}\\\hline\end{array}$

Vậy $x∈\big\{\dfrac{1}{9};\dfrac{64}{9}\big\}$ thì $\dfrac{7P}{3}∈\Bbb Z$