Đáp án:
Giải thích các bước giải:
`P=(\frac{1}{\sqrt{x}+1}-\frac{2}{x-1}):(\frac{2}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+1})`
ĐK: `x \ge 0, x \ne 1`
a) `P=[\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}]:[\frac{2(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}]`
`P=\frac{\sqrt{x}-1-2}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{2\sqrt{x}+2-\sqrt{x}+1}`
`P=\frac{\sqrt{x}-3}{(\sqrt{x}-1)(\sqrt{x}+1)}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}+3}`
`P=\frac{\sqrt{x}-3}{\sqrt{x}+3}`
b) `x=4-2\sqrt{3}`
`x=4+1-2\sqrt{3}`
`x=(\sqrt{3}-1)^2`
`⇒ \sqrt{x}=\sqrt{(\sqrt{3}-1)^2}=|\sqrt{3}-1|=\sqrt{3}-1`
Thay vào P ta có:
`P=\frac{\sqrt{3}-1-3}{\sqrt{3}-1+3}=\frac{\sqrt{3}-4}{\sqrt{3}+2}=6\sqrt{3}-11`
Vậy khi `x=4-2\sqrt{3}` thì `P=6\sqrt{3}-11`
c) `P \le 0`
`⇔ \frac{\sqrt{x}-3}{\sqrt{x}+3} \le 0`
Ta có: `x \ge 0⇒\sqrt{x} \ge 0⇒ \sqrt{x}+3 \ge 3`
`⇔ \sqrt{x}-3 \le 0`
`⇔ \sqrt{x} \le 3`
`⇔ x \le 9` kết hợp ĐKXĐ
Vậy `0 \le x \le 9, x \ne 1` thì `P \le 0`
