$(x>0)$
`\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.
