`\color{#FF0000}{T}\color{#FF7F24}{i}\color{#FFFF00}{t}\color{#7FFF00}{u}\color{#00F5FF}{u}\color{#FF1493}{u}`

$3)$ Có `A.B=(3\sqrt{x})/(\sqrt{x}-3).(\sqrt{x}-3)/(\sqrt{x}+1) (x>=0; x!=9)`

`=(3\sqrt{x})/(\sqrt{x}+1)`

`=(3(\sqrt{x}+1)-3)/(\sqrt{x}+1)`

`=3-(3)/(\sqrt{x}+1)`

`-> A.B` là STN khi `3` $\vdots$ `\sqrt{x}+1` và `(3)/(\sqrt{x}+1)<=3` (luôn đúng `AA x>=0; x!=9; x in ZZ`)

`-> \sqrt{x}+1 in Ư_3={-3; -1; 1; 3}` 

Mà `\sqrt{x}+1>=1>0 AA x>=0; x!=9; x in ZZ`

`-> \sqrt{x}+1 in {1; 3}`

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

`-> x in {0; 4} (TM)`

Vậy `x in {0; 4}`