`***`Lời giải`***`
a)
`\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}`
`=\frac{(\sqrt{3}-\sqrt{6})(1+\sqrt{2})-(2+\sqrt{8})(1-\sqrt{2})}{1-2}`
`=-[(\sqrt{3}-\sqrt{6})(1+\sqrt{2})-(2+\sqrt{8})(1-\sqrt{2})]`
`=-(2-\sqrt{3})`
`=\sqrt{3}-2`
b)
`B=(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}-1}{\sqrt{x}-1}):\frac{\sqrt{x}}{x+\sqrt{x}}`
ĐKXĐ: `x>0`
`=(\frac{1}{\sqrt{x}}+1):\frac{\sqrt{x}}{x+\sqrt{x}}`
`=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}}{x+\sqrt{x}}`
`=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}+1)}{x}`
`=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}`
Vậy` B=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}` với `x>0`
