Đáp án:
Giải thích các bước giải:
`P=(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}):(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1})`
a) `ĐKXĐ: x >0, x \ne 1`
`P=[\frac{x}{\sqrt{x}(\sqrt{x}-1)}-\frac{1}{\sqrt{x}(\sqrt{x}-1)}]:[\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}]`
`P=[\frac{x-1}{\sqrt{x}(\sqrt{x}-1)}]:[\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}]`
`P=[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}].[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}+1}]`
`P=\frac{x-1}{\sqrt{x}}`
b) `P>0`
`⇔ \frac{x-1}{\sqrt{x}}>0`
Vì `\sqrt{x}>0`
`⇒ x-1>0`
`⇔ x>1` kết hợp ĐKXĐ
Vậy `x>1` thì `P>0`
c) `P=6`
`⇔ frac{x-1}{\sqrt{x}}=6`
`⇔ x-1=6\sqrt{x}`
`⇔` \(\left[ \begin{array}{l}x=3+\sqrt{10}\ (TM)\\x=3-\sqrt{10}\ (Loại)\end{array} \right.\)
Vậy `x=3+\sqrt{10}` thì `P=6`
