ĐKXĐ: `x\geq0,x\ne1`

`a,` `x=9` (thỏa mãn ĐKXĐ)

Thay `x=9` vào `A` có:

`A={\sqrt{9}+2}/{\sqrt{9}-1}={3+2}/{3-1}=5/2`

Vậy với `x=9` thì `A=5/2`

`b,` 

`B={\sqrt{x}}/{\sqrt{x}-1}+{5}/{\sqrt{x}+1}-{8\sqrt{x}-6}/{x-1}={\sqrt{x}}/{\sqrt{x}-1}+{5}/{\sqrt{x}+1}-{8\sqrt{x}-6}/{(\sqrt{x}-1)(\sqrt{x}+1)}={\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}+{5(\sqrt{x}-1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}-{8\sqrt{x}-6}/{(\sqrt{x}-1)(\sqrt{x}+1)}={\sqrt{x}(\sqrt{x}+1)+5(\sqrt{x}-1)-(8\sqrt{x}-6)}/{(\sqrt{x}-1)(\sqrt{x}+1)}={x+\sqrt{x}+5\sqrt{x}-5-8\sqrt{x}+6}/{(\sqrt{x}-1)(\sqrt{x}+1)}={x-2\sqrt{x}+1}/{(\sqrt{x}-1)(\sqrt{x}+1)}={(\sqrt{x}-1)^2}/{(\sqrt{x}-1)(\sqrt{x}+1)}={\sqrt{x}-1}/{\sqrt{x}+1}`

Vậy với `x\geq0,x\ne1` thì `B={\sqrt{x}-1}/{\sqrt{x}+1}`

`c,`

`A.B={\sqrt{x}+2}/{\sqrt{x}-1}.{\sqrt{x}-1}/{\sqrt{x}+1}={\sqrt{x}+2}/{\sqrt{x}+1}`

`A.B={\sqrt{x}}/{2}`

`⇔{\sqrt{x}+2}/{\sqrt{x}+1}={\sqrt{x}}/{2}`

`⇔{\sqrt{x}+2}/{\sqrt{x}+1}-{\sqrt{x}}/{2}=0`

`⇔{2(\sqrt{x}+2)}/{2(\sqrt{x}+1)}-{\sqrt{x}(\sqrt{x}+1)}/{2(\sqrt{x}+1)}=0`

`⇔{2(\sqrt{x}+2)-\sqrt{x}(\sqrt{x}+1)}/{2(\sqrt{x}+1)}=0`

`⇔2(\sqrt{x}+2)-\sqrt{x}(\sqrt{x}+1)=0`

`⇔2\sqrt{x}+4-x-\sqrt{x}=0`

`⇔-x+\sqrt{x}+4=0`

`⇔x-\sqrt{x}-4=0`

`⇔x-\sqrt{x}-4=0`

`⇔x-\sqrt{x}+{1}/{4}-{17}/{4}=0`

`⇔(\sqrt{x}-{1}/{2})^2={17}/{4}`

⇔\(\left[ \begin{array}{l}\sqrt{x}-\frac{1}{2}=\frac{\sqrt{17}}{2}(1)\\\sqrt{x}-\frac{1}{2}=-\frac{\sqrt{17}}{2}(2)\end{array} \right.\)

Giải `(1)`

`\sqrt{x}-\frac{1}{2}=\frac{\sqrt{17}}{2}`

`⇔\sqrt{x}=\frac{1+\sqrt{17}}{2}`

`⇔x=\frac{9+\sqrt{17}}{2}` (Thỏa mãn)

Giải `(2)`

`\sqrt{x}-\frac{1}{2}=-\frac{\sqrt{17}}{2}`

`⇔\sqrt{x}=\frac{1-\sqrt{17}}{2}` (vô lí vì `\sqrt{x}\geq0` với mọi `x\geq0`, `\frac{1-\sqrt{17}}{2}<0`)

Vậy với `x=\frac{9+\sqrt{17}}{2}` thì `A.B={\sqrt{x}}/{2}`