`B=\frac{\sqrt{x}+\sqrt{y}}{2\sqrt{x}-2\sqrt{y}}-\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+2\sqrt{y}}-\frac{x+y}{y-x}(x,y>=0)`
`=\frac{\sqrt{x}+\sqrt{y}}{2(\sqrt{x}-\sqrt{y})}-\frac{\sqrt{x}-\sqrt{y}}{2(\sqrt{x}+\sqrt{y})}+\frac{x+y}{x-y}`
`=\frac{(\sqrt{x}+\sqrt{y})^2-(\sqrt{x}-\sqrt{y})^2+2x+2y}{2(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}`
`=\frac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}`
`=\frac{2(x+2\sqrt{xy}+y)}{2(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}`
`=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}`
Tại `x=16` và `y=5`
`=\frac{\sqrt{16}+\sqrt{5}}{\sqrt{16}-\sqrt{5}}`
`=\frac{4+\sqrt{5}}{4-\sqrt{5}}`
`=\frac{(4+\sqrt{5})^2}{16-5}`
`=\frac{16+8\sqrt{5}+5}{11}`
`=\frac{21+8\sqrt{5}}{11}`
Vậy tại `x=16` và `y=5`
thì `B=\frac{21+8\sqrt{5}}{11}`
