Question

Cho P = \(\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\), tìm x để P nguyên.

Answer 1

\(đkx\ge0\\ P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\\ =1+\dfrac{5}{\sqrt{x}+2}\)

=>  \(\sqrt{x}+2\inƯ\left(5\right)\)

mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)

=> \(x=9\left(thoamanđk\right)\)

Vậy \(x=9\) thì P nguyên 

Answer 2

`P = 1 + 5/(sqrt x + 2)`

Nếu `sqrt x in II -> 5/(sqrt x + 2)` không thể nguyên.

`-> sqrt x in ZZ -> x` chính phương.

`-> 5/(sqrt x + 2) in ZZ`.

`-> 5 vdots sqrt x + 2`

`-> sqrt x + 2 in Ư(5)`

`-> sqrt x in {-1, -3, 3, -7}`

`-> x = 9`.

Answer 3

`P=[\sqrt{x}+7]/[\sqrt{x}+2]`

`P=[\sqrt{x}+2+5]/[\sqrt{x}+2]`

`P=1+5/[\sqrt{x}+2]`

Để `P in ZZ<=>1+5/[\sqrt{x}+2] in ZZ`

    `=>5/[\sqrt{x}+2] in ZZ`

   `=>\sqrt{x}+2 in Ư_5`

Mà `Ư_5={+-1;+-5}`

`@\sqrt{x}+2=1=>\sqrt{x}=-1` (Vô lí)

`@\sqrt{x}+2=-1=>\sqrt{x}=-3` (Vô lí)

`@\sqrt{x}+2=5=>\sqrt{x}=3=>x=9` (t/m)

`@\sqrt{x}+2=-5=>\sqrt{x}=-7` (Vô lí)

 Vậy `x=9`