`B=(1-\frac{4\sqrt{x}}{x-1}+\frac{1}{\sqrt{x}-1}):\frac{x-2\sqrt{x}}{x-1}(x>=0,x\ne1;x\ne4)`
`=(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}).\frac{x-1}{\sqrt{x}(\sqrt{x}-2)}`
`=\frac{x-3\sqrt{x}}{x-1}.\frac{x-1}{\sqrt{x}(\sqrt{x}-2)}`
`=\frac{\sqrt{x}-3}{\sqrt{x}-2}`
`b)x=11-6\sqrt{2}`
`->\sqrt{x}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}=\sqrt{(3-\sqrt{2})^2}=3-\sqrt{2}`
Thế `\sqrt{x}=3-\sqrt{2}` vào `B`, ta có:
`B=\frac{3-\sqrt{2}-3}{3-\sqrt{2}-2}`
`=\frac{-\sqrt{2}}{1-\sqrt{2}}`
`=\frac{-\sqrt{2}(1+\sqrt{2})}{1-2}`
`=\sqrt{2}(1+\sqrt{2})`
`=\sqrt{2}+2`
Vậy `x=11-6\sqrt{2}` thì `B=2+\sqrt{2}`
