Đáp án:
Giải thích các bước giải:
1) `A=\frac{x+2\sqrt{x}}{x-1}\ (x \ge 0, x \ne 1)`
Thay `x=25` vào BT A ta có:
`A=\frac{25+2\sqrt{25}}{25-1}=\frac{25+2.5}{24}=35/24`
Vậy với `x=25` thì `A=35/24`
2) `B=\frac{2}{x}-\frac{2-x}{x(\sqrt{x}+1)}`
ĐK: `x>0`
`B=\frac{2(\sqrt{x}+1)}{x(\sqrt{x}+1)}-\frac{2-x}{x(\sqrt{x}+1)}`
`B=\frac{2\sqrt{x}+2-2+x}{x(\sqrt{x}+1)}`
`B=\frac{2\sqrt{x}+x}{x(\sqrt{x}+1)}`
`B=\frac{\sqrt{x}(2+\sqrt{x})}{\sqrt{x}.\sqrt{x}(\sqrt{x}+1)}`
`B=\frac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}`
3) Đặt `P=\frac{A}{B}`
`P=\frac{x+2\sqrt{x}}{x-1}:\frac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}+1)}`
ĐK: `x >0, x \ne 1`
`P=\frac{\sqrt{x}(\sqrt{x}+2)}{(\sqrt{x}+1)(\sqrt{x}-1)}.\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}+2}`
`P=\frac{x}{\sqrt{x}-1}`
`P>1`
`⇔ \frac{x}{\sqrt{x}-1}>1`
`⇔ \frac{x}{\sqrt{x}-1}-1>0`
`⇔ \frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0`
Ta có: `x-\sqrt{x}+1=(\sqrt{x})^2-2.\frac{1}{2}.\sqrt{x}+(\frac{1}{2})^2+3/4`
`=(\sqrt{x}+1/2)^2+3/4`
Do `x>0⇒ \sqrt{x} >0 ⇒ (\sqrt{x}+1/2)^2+3/4 \ge 3/4 ∀x`
`⇒ \sqrt{x}-1>0`
`⇔ \sqrt{x}>1`
`⇔ x>1` kết hợp ĐKXĐ
Vậy `x>1` thì `P>1`
