Đáp án:

$x = 9$ $thì$ $A∈Z$

Giải thích các bước giải:

$ĐK : x > 0$

$A = \frac{3√x + 5}{2√x+1}$

$2A = \frac{6√x + 10}{2√x + 1}$

$2A = \frac{3(2√x + 1) + 7}{2√x + 1}$

$2A = 3 + \frac{7}{2√x + 1}$

$Để$ $A ∈ Z$$→2A ∈Z$

$→\frac{7}{2√x + 1}∈Z$

$→ 7 ⋮2√x + 1$

$→2√x + 1 ∈Ư(7)$

$Vì$ $x > 0→√x > 0 →2√x > 0 →2√x + 1 > 1$

$→ 2√x + 1 = 7 $

$→2√x = 6$

$→√x = 3$

$→ x = 9 (tmđk)$

$Vậy$ $x = 9$ $thì$ $A∈Z$

$(x>0)$

`\qquad A={3\sqrt{x}+5}/{2\sqrt{x}+1}\in Z`

Vì `2\in Z; A\in Z=>2A\in Z`

`=>{2(3\sqrt{x}+5)}/{2\sqrt{x}+1}\in Z`

`=>{6\sqrt{x}+10}/{2\sqrt{x}+1}\in Z`

`=>{3(\2\sqrt{x}+1)+7}/{2\sqrt{x}+1}\in Z`

`=>{3(2\sqrt{x}+1)}/{2\sqrt{x}+1}+7/{2\sqrt{x}+1}\in Z`

`=>3+ 7/{2\sqrt{x}+1}\in Z`

`=>7/{2\sqrt{x}+1}\in Z`

`=>2\sqrt{x}+1 \in Ư(7)={-7;-1;1;7}`

`=>2\sqrt{x}\in {-8;-2;0;6}`

`=>\sqrt{x}\in {-4;-1;0;3}`

Vì `\sqrt{x}> 0` với mọi $x>0$

`=>\sqrt{x}=3`

`<=>x=9`

Vậy $x=9$ thỏa đề bài.