Tìm x để p là STN P=`(2\sqrt{x}+7)/(\sqrt{x}+1)`

`P = (2sqrtx  +7)/(sqrtx +1) (x >= 0)`

`= (2(sqrtx + 1) + 5)/(sqrtx + 1)`

`= 2 + 5/(sqrtx + 1)`

`sqrtx >= 0`

`=> sqrtx + 1 >= 0 + 1 =1`

`=> 5/(sqrtx + 1) <= 5/1 = 5`

`=> P = 2 + 5/(sqrtx + 1)`

`<= 2 + 5 = 7`

Mà `2sqrtx + 7 > 0 ; sqrtx + 1 > 0`

`=> P =(2sqrtx + 7)/(sqrtx + 1) > 0`

`=> 0 < P <= 7`

Xét `P = 1`

`=> (2sqrtx + 7)/(sqrtx +1) = 1`

`<=> 2sqrtx + 7 = sqrtx + 1`

`<=> sqrtx = 6`

`<=> x  = 36(TM)`

Xét `P = 2`

`=> (2sqrtx + 7)/(sqrtx + 1) = 2`

`<=> 2sqrtx + 7 = 2sqrtx + 2`

`<=> 7 = 2 (l)`

Xét `P = 3`

`=> (2sqrtx + 7)/(sqrtx + 1) = 3`

`<=> 2sqrtx + 7 = 3sqrtx + 3`

`<=> sqrtx = 4 `

`<=> x = 16 (TM)`

Xét `P = 4`

`=> (2sqrtx + 7)/(sqrtx + 1) = 4`

`<=> 2sqrtx + 7 = 4sqrtx + 4`

`<=> 2sqrtx = 3`

`<=> sqrtx = 3/2`

`<=> x = 9/4 (TM)`

Xét `P = 5`

`=> (2sqrtx + 7)/(sqrtx + 1) =5`

`<=>2sqrtx + 7 = 5sqrtx + 5`

`<=> 3sqrtx = 2`

`<=> sqrtx = 2/3`

`<=>x = 4/9 (TM)`

Xét `P = 6`

`=> (2sqrtx + 7)/(sqrtx + 1) =6`

`<=> 2sqrtx + 7 = 6sqrtx + 6`

`<=> 4sqrtx = 1`

`<=> sqrtx = 1/4`

`<=> x = 1/16 (TM)`

Xét `P =7`

`=> (2sqrtx + 7)/(sqrtx + 1) = 7`

`<=> 2sqrtx + 7 = 7sqrtx + 7`

`<=> 5sqrtx =0`

`<=> sqrtx = 0`

`<=>x =0 (TM)`
Vậy `x = {0 ; 1/16 ; 4/9 ; 9/4 ; 16 ; 36}` thì `P \in NN`