`\color{#274e13}{♡}\color{#38761d}{C}\color{#55953a}{h}\color{#79b95d}{a}\color{#98d97c}{n}\color{#b7fa99}{h}`

`đk: x > 0, x ne 1`

`P = ((sqrtx + 2)/(x + 2 sqrtx + 1) - (sqrtx - 2)/(sqrtx - 1)) . (sqrtx + 1)/sqrtx`

`= ((sqrtx + 2)/((sqrtx + 1)^2) - (sqrtx - 2)/(sqrtx - 1)) . (sqrtx + 1)/sqrtx`

`= (((sqrtx + 2)(sqrtx - 1) - (sqrtx - 2)(sqrtx + 1)^2)/((sqrtx + 1)^2 . (sqrtx - 1))) . (sqrtx + 1)/sqrtx`

`= ((x + sqrtx - 2 - (sqrtx - 2)(x + 2 sqrtx + 1))/((sqrtx + 1)^2 . (sqrtx - 1))) . (sqrtx + 1)/sqrtx`

`= ((x + sqrtx - 2 - (x sqrtx + 2x + sqrtx - 2x - 4 sqrtx - 2))/((sqrtx + 1)^2 . (sqrtx - 1))) . (sqrtx + 1)/sqrtx`

`= ((-x sqrtx + x + 4 sqrtx)/((sqrtx + 1)^2 . (sqrtx - 1))) . (sqrtx + 1)/sqrtx`

`= (sqrtx (-x + sqrtx + 4)/((sqrtx + 1)^2 . (sqrtx - 1))) . (sqrtx + 1)/sqrtx`

`= (-x + sqrtx + 4)/((sqrtx + 1)(sqrtx - 1))`

`= (-x + sqrtx + 4)/(x - 1)`

`= -(x - 1 - sqrtx - 3)/(x - 1)`

`= -1 + (sqrtx + 3)/(x - 1)`

`= -1 + (sqrtx + 3)/((sqrtx - 1)(sqrtx + 1))`

Để `P in ZZ to (sqrtx + 3)/(x - 1) in ZZ`

Đặt `a = sqrtx (a > 0, a ne 1)`

`(a + 3)/(a^2 - 1) in ZZ`

`(a + 3) vdots (a^2 - 1)`

`(a + 3)(a - 3) vdots (a^2 - 1)`

`(a^2 - 9) vdots (a^2 - 1)`

`(a^2 - 1 - 8) vdots (a^2 - 1)`

`8 vdots (a^2 - 1)`

`a^2 - 1 in Ư(8) = {-8; -4; -2; -1; 1; 2; 4; 8}`

`to a^2 in {-7; -3; -1; 0; 2; 3; 5; 9}`

Vì `a > 0` và `a ne 1` nên `a^2 > 0` và `a^2 ne 1`

`to a^2 in {2; 3; 5; 9}`

`to x in {2; 3; 5; 9}`

Thử lại `:`

`x = 2 to P = -1 + (sqrt2 + 3)/1 = 2 + sqrt2  (loại)`

`x = 3 to P = -1 + (sqrt3 + 3)/2  (loại)`

`x = 5 to P = -1 + (sqrt5 + 3)/4  (loại)`

`x = 9 to P = -1 + (3 + 3)/8 = -1 + 6/8 = -1/4 (loại)`

Vậy không có giá trị nào của `x` thỏa mãn để `P in ZZ.`