Đáp án  + Giải thích các bước giải:

 1) Thay $x = 9$ vào $A = \dfrac{3\sqrt{x} - 1}{\sqrt{x} + 1}$, ta có:

$A = \dfrac{3\sqrt{9} - 1}{\sqrt{9} + 1} = \dfrac{3 . 3 - 1}{3 + 1} = \dfrac{8}{4} = 2$

Vậy với $x = 9$ thì $A = 2$

2) $B = \dfrac{2\sqrt{x}}{\sqrt{x} + 3} + \dfrac{4}{\sqrt{x} - 3} + \dfrac{x + 15}{9 - x}$

$\Leftrightarrow B = \dfrac{2\sqrt{x}}{\sqrt{x} + 3} + \dfrac{4}{\sqrt{x} - 3} - \dfrac{x + 15}{x - 9}$

$\Leftrightarrow B = \dfrac{2\sqrt{x}}{\sqrt{x} + 3} + \dfrac{4}{\sqrt{x} - 3} - \dfrac{x + 15}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{2\sqrt{x}(\sqrt{x} - 3)}{(\sqrt{x} + 3)(\sqrt{x} - 3)} + \dfrac{4(\sqrt{x} + 3)}{(\sqrt{x} + 3)(\sqrt{x} - 3)} - \dfrac{x + 15}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{2x - 6\sqrt{x}}{(\sqrt{x} + 3)(\sqrt{x} - 3)} + \dfrac{4\sqrt{x} + 12}{(\sqrt{x} + 3)(\sqrt{x} - 3)} - \dfrac{x + 15}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{2x - 6\sqrt{x} + 4\sqrt{x} + 12 - x - 15}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{x - 2\sqrt{x} - 3}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{(\sqrt{x} + 1)(\sqrt{x} - 3)}{(\sqrt{x} + 3)(\sqrt{x} - 3)}$

$\Leftrightarrow B = \dfrac{\sqrt{x} + 1}{\sqrt{x} + 3}$

Thay $B = \dfrac{1}{2}$, ta có:

$\dfrac{\sqrt{x} + 1}{\sqrt{x} + 3} = \dfrac{1}{2}$

$\Leftrightarrow 2(\sqrt{x} + 1) = \sqrt{x} + 3$
$\Leftrightarrow 2\sqrt{x} + 2 = \sqrt{x} + 3$

$\Leftrightarrow 2\sqrt{x} - \sqrt{x} = -2 + 3$
$\Leftrightarrow \sqrt{x} = 1$

$\Leftrightarrow x = 1$
Vậy với $x = 1$ thì $B = \dfrac{1}{2}$

3) Ta có: $Q = A . B$

$\Rightarrow Q = \dfrac{3\sqrt{x} - 1}{\sqrt{x} + 1} . \dfrac{\sqrt{x} + 1}{\sqrt{x} + 3}$

$\Leftrightarrow Q = \dfrac{3\sqrt{x} - 1}{\sqrt{x} + 3}$

Để $Q$ là một số tự nhiên thì $3\sqrt{x} - 1$`\vdots`$\sqrt{x} + 3$

$\Rightarrow 3\sqrt{x} + 9 - 10$`\vdots`$\sqrt{x} + 3$

$\Leftrightarrow 3(\sqrt{x} + 3) - 10$`\vdots`$\sqrt{x} + 3$
$\Leftrightarrow -10$`\vdots`$\sqrt{x} + 3$

$\Leftrightarrow \sqrt{x} + 3 \in$`{-10; -5; -2; -1; 1; 2; 5; 10}`

$\Leftrightarrow \sqrt{x} \in$`{-13; -8; -5; -4; -2; -1; 2; 7}`

Mà $\sqrt{x} \ge 0$

$\Rightarrow \sqrt{x} \in$`{2; 7}`

$\Rightarrow x \in$`{4; 49}`

Vậy với $x \in$`{4; 49}` thì giá trị của $Q$ là một số tự nhiên

`I)`

`A=(3sqrtx-1)/(sqrtx+1)`

`B=(2sqrtx)/(sqrtx+3)+(4)/(sqrtx-3)+(x+15)/(9-x)` 

`xge0` `xne9`

`1)`

Thay `x=9` (tm đkxđ) vào A ta được :

`A=(3sqrtx-1)/(sqrtx+1)` `=` `(3sqrt9-1)/(sqrt9+1)``=``(3.3-1)/(3+1)``=``(8)/(4)``=``2`

`2)`

`B=(2sqrtx)/(sqrtx+3)+(4)/(sqrtx-3)+(x+15)/(9-x)` 

`=(2sqrtx(sqrtx-3))/((sqrtx+3)(sqrtx-3))+(4(sqrtx+3))/((sqrtx-3)(sqrtx+3))-(x+15)/(x-9)` 

`=(2x-6sqrtx)/((sqrtx+3)(sqrtx-3))+(4sqrtx+12)/((sqrtx-3)(sqrtx+3))-(x+15)/((sqrtx-3)(sqrtx+3))`

`=(2x-6sqrtx+4sqrtx+12-x-15)/((sqrtx+3)(sqrtx-3))`

`=(x-2sqrtx-3)/((sqrtx+3)(sqrtx-3))`

`=(x+sqrtx-3sqrtx-3)/((sqrtx+3)(sqrtx-3))`

`=(sqrtx(sqrtx+1)-3(sqrtx+1))/((sqrtx+3)(sqrtx-3))`

`=((sqrtx-3)(sqrtx+1))/((sqrtx+3)(sqrtx-3))`

`=(sqrtx+1)/(sqrtx+3)`

Để `B=1/2` thì `(sqrtx+1)/(sqrtx+3)=1/2`

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

`⇔``2(sqrtx+1)=sqrtx+3`

`⇔``2sqrtx+2=sqrtx+3`

`⇔``2sqrtx-sqrtx=3-2`

`⇔``sqrtx=1`

`⇔``x=1` `(tm)`

Vậy để `B=1/2` thì `x=1`

`3)`

Ta có: `Q = A . B`
`\Rightarrow Q = \frac{3\sqrt{x} - 1}{\sqrt{x} + 1} . \frac{\sqrt{x} + 1}{\sqrt{x} + 3}`
`\Leftrightarrow Q = \frac{3\sqrt{x} - 1}{\sqrt{x} + 3}`

Để `Q` là một số tự nhiên thì `3\sqrt{x} - 1``\vdots``\sqrt{x} + 3` và `(3sqrtx-1)/(sqrtx+3)ge0`

`(3sqrtx-1)/(sqrtx+3)ge0`

`⇔` `3\sqrt{x} - 1ge0`

`⇔` `3\sqrt{x} - 1ge0`

`⇔``3\sqrt{x}ge1`

`⇔``\sqrt{x}ge1/3`

`⇔``xge1/9`

`3\sqrt{x} - 1``\vdots``\sqrt{x} + 3`

$\Leftrightarrow 3(\sqrt{x} + 3) - 10$`\vdots`$\sqrt{x} + 3$
$\Leftrightarrow -10$`\vdots`$\sqrt{x} + 3$

$\Leftrightarrow \sqrt{x} + 3 \in$`{-10; -5; -2; -1; 1; 2; 5; 10}`

`sqrtx+3=-10`

`⇔` `sqrtx=-13(ktm)`

`sqrtx+3=-5`

`⇔` `sqrtx=-8(ktm)`

`sqrtx+3=-2`

`⇔` `sqrtx=-5(ktm)`

`sqrtx+3=-1`

`⇔` `sqrtx=-4(ktm)`

`sqrtx+3=1`

`⇔` `sqrtx=-2(ktm)`

`sqrtx+3=2`

`⇔` `sqrtx=-1(ktm)`

`sqrtx+3=5`

`⇔` `sqrtx=2`

`⇔` `x=4(tm)`

`sqrtx+3=10`

`⇔` `sqrtx=7`

`⇔` `x=49(tm)`

$\Rightarrow x \in$`{4; 49}`

Vậy với `x in{4; 49}` thì Q là một số tự nhiên