`a.`
`B= \frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{(2-\sqrt{x})(2+\sqrt{x})}`
`= \frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}-\frac{2+5\sqrt{x}}{(\sqrt{x}-2)(2+\sqrt{x})}`
`= \frac{(\sqrt{x}+2)(\sqrt{x}+1)+2\sqrt{x}(\sqrt{x}-2)-(2+5\sqrt{x})}{(\sqrt{x}-2)(2+\sqrt{x})}`
`= \frac{x+\sqrt{x}+2\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{((\sqrt{x}-2)(\sqrt{x}+2)}`
`= \frac{3x-6\sqrt{x}}{(\sqrt{x-2})(\sqrt{x}+2)}`
`= \frac{3\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}`
`= \frac{3\sqrt{x}}{\sqrt{x}+2}`
`b.`
Để `B=2` thì `\frac{3\sqrt{x}}{\sqrt{x}+2}=2`
`<=> 3\sqrt{x}=2(\sqrt{x}+2)`
`<=> 3\sqrt{x}=2\sqrt{x}+4`
`<=> 3\sqrt{x}-2\sqrt{x}=4`
`<=> \sqrt{x}=4`
`<=> x=16`(tm)
Vậy `x=16` thì `B=2`
