`***`Lời giải`***`
2)
a)
ĐKXĐ: `x>0;xne1`
`P=(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}).\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=(\frac{x-2}{\sqrt{x}(\sqrt{x}+2)}+\frac{1}{\sqrt{x}+2}).\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{x+\sqrt{x}-2}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{(\sqrt{x}+2)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+2)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{\sqrt{x}+1}{\sqrt{x}}`
b)
Ta có: `2P=2\sqrt{x}+5`
`=>\frac{2(\sqrt{x}+1)}{\sqrt{x}}=2\sqrt{x}+5`
`<=>\frac{2\sqrt{x}+2}{\sqrt{x}}=2\sqrt{x}+5`
`<=>\frac{2\sqrt{x}+2}{\sqrt{x}}-(2\sqrt{x}+5)=0`
`<=>\frac{2\sqrt{x}+2-\sqrt{x}(2\sqrt{x}+5)}{\sqrt{x}}=0`
`<=>\frac{2\sqrt{x}+2-\sqrt{x}(2\sqrt{x}+5)}{\sqrt{x}}=0`
`<=>\frac{2\sqrt{x}+2-2x-5\sqrt{x}}{\sqrt{x}}=0`
`<=>\frac{-2x-3\sqrt{x}+2}{\sqrt{x}}=0`
`=>-2x-\sqrt{x}+4\sqrt{x}+2=0`
`<=>-\sqrt{x}(2\sqrt{x}+1)+2(2\sqrt{x}+1)=0`
`<=>(2\sqrt{x}+1)(2-\sqrt{x})=0`
`<=>`\(\left[ \begin{array}{l}2\sqrt{x}+1=0\\2-\sqrt{x}=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2\sqrt{x}=-1\\-\sqrt{x}=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}\sqrt{x}=\dfrac{-1}{2}\\\sqrt{x}=2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{4}(N)\\x=4(N)\end{array} \right.\)
Vậy `S={\frac{1}{4};4}`
