Đáp án:
Giải thích các bước giải:
`N=(\frac{\sqrt{x}}{2+\sqrt{x}}+\frac{x+4}{4-x}):(\frac{1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x-2\sqrt{x}})`
ĐK: `x > 0, x \ne 4`
`N=[\frac{\sqrt{x}(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}+\frac{x+4}{(2-\sqrt{x})(2+\sqrt{x})}]:[\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-2)}+\frac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-2)}]`
`N=[\frac{2\sqrt{x}-x+x+4}{(2+\sqrt{x})(2-\sqrt{x})}]:[\frac{\sqrt{x}-2+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-2)}]`
`N=[\frac{2\sqrt{x}+4}{(2+\sqrt{x})(2-\sqrt{x})}].[\frac{\sqrt{x}(\sqrt{x}-2)}{3\sqrt{x}-1}]`
`N=[\frac{2(\sqrt{x}+2)}{(2+\sqrt{x})(2-\sqrt{x})}].[\frac{\sqrt{x}(\sqrt{x}-2)}{3\sqrt{x}-1}]`
`N=\frac{-2\sqrt{x}}{3\sqrt{x}-1}`
b) `N=1`
`⇔\frac{-2\sqrt{x}}{3\sqrt{x}-1}=1`
`⇔ 3\sqrt{x}-1=-2\sqrt{x}`
`⇔ 5\sqrt{x}=1`
`⇔ x=1/25\ (TM)`
Vậy `x=1/25` thì `N=1`
