`\frac{1}{\sqrt{x-\sqrt{2x}-1}}`
`=\frac{1}{\sqrt{x-2.\sqrt{x} . \frac{1}{\sqrt{2}} +1/2 -3/2}}`
`=\frac{1}{\sqrt{(x-\frac{1}{\sqrt{2})^2-3/2)}`
ĐKXĐ: `(x-\frac{1}{\sqrt{2}})^2-3/2>0`
`x>=0`
`<=>(x-1/(\sqrt{2})-\frac{\sqrt{3}}{\sqrt{2}})(x-1/(\sqrt{2})+\frac{\sqrt{3}}{\sqrt{2}])>0`
`<=>[({(x-\frac{1+\sqrt{3}}{\sqrt{2}}>0),(x-\frac{1-\sqrt{3}}{\sqrt{2}}>0):}),({(x-\frac{1+\sqrt{3}}{\sqrt{2}}<0),(x-\frac{1-\sqrt{3}}{\sqrt{2}}<0):}):}`
`<=>[({(x>\frac{1+\sqrt{3}}{\sqrt{2}}),(x>\frac{1-\sqrt{3}}{\sqrt{2}}):}),({(x<\frac{1+\sqrt{3}}{\sqrt{2}}),(x<\frac{1-\sqrt{3}}{\sqrt{2}}):}):}`
`=>[(x>\frac{1+\sqrt{3}}{\sqrt{2}}),(x<\frac{1-\sqrt{3}}{\sqrt{2}}):}`
Mà `x>=0`
`=>x>\frac{1+\sqrt{3}}{\sqrt{2}}`
Vậy ta có ĐKXĐ: `x>\frac{1+\sqrt{3}}{\sqrt{2}}`
