`***`Lời giải`***`
a)
ĐKXĐ: `x≥0;xne±1`
`P=(\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}):(1+\frac{\sqrt{x}}{x+1})`
`=(\frac{2\sqrt{x}}{x(\sqrt{x}-1)+\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}):(1+\frac{\sqrt{x}}{x+1})`
`=\frac{2\sqrt{x}-(x+1)}{(\sqrt{x}-1)(x+1)}:\frac{x+1+\sqrt{x}}{x+1}`
`=\frac{-x+2\sqrt{x}-1}{(\sqrt{x}-1)(x+1)}:\frac{x+1+\sqrt{x}}{x+1}`
`=\frac{-\sqrt{x}(\sqrt{x}-1)+\sqrt{x}-1}{(\sqrt{x}-1)(x+1)}:\frac{x+1+\sqrt{x}}{x+1}`
`=\frac{-(\sqrt{x}-1)^2}{(\sqrt{x}-1)(x+1)}.\frac{x+1}{x+1+\sqrt{x} }`
`=\frac{-\sqrt{x}+1}{x+\sqrt{x}+1 }`
b)
Ta có:` P≤0`
`=>\frac{-\sqrt{x}+1}{x+\sqrt{x}+1 }≤0`
Ta lại có:`x+\sqrt{x}+1=x+\sqrt{x}+1/4+3/4=(\sqrt{x}+1/2)^2+3/4`
Mà `(\sqrt{x}+1/2)^2≥0` với `∀x`
`<=>(\sqrt{x}+1/2)^2+3/4≥3/4>0`
`=>-\sqrt{x}+1≤0`
`<=>-\sqrt{x}≤-1`
`<=>\sqrt{x}≥1`
`<=>x≥1`
Kết hợp với ĐKXĐ:`x≥0;xne±1`
Vậy `x>1` thì ` P≤0`
