Đáp án:
Giải thích các bước giải:
1) `P=(\frac{x+6}{\sqrt{x}-1}-\sqrt{x})(2-\frac{\sqrt{x}+3}{\sqrt{x}+1})`
ĐK: `x \ge 0, x \ne 1`
`P=[\frac{x+6}{\sqrt{x}-1}-\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}][\frac{2(\sqrt{x}+1)}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}+1}]`
`P=(\frac{x+6-x+\sqrt{x}}{\sqrt{x}-1})(\frac{2\sqrt{x}+2-\sqrt{x}-3}{\sqrt{x}+1})`
`P=\frac{\sqrt{x}+6}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`P=\frac{\sqrt{x}+6}{\sqrt{x}+1}`
5)
`3P=2\sqrt{x}+4`
`⇔ \frac{3(\sqrt{x}+6)}{\sqrt{x}+1}=2\sqrt{x}+4`
`⇔ (2\sqrt{x}+4)(\sqrt{x}+1)=3(\sqrt{x}+6)`
`⇔ 2x+6\sqrt{x}+4=3\sqrt{x}+18`
`⇔ 2x+3\sqrt{x}-14=0`
`⇔ (\sqrt{x}-2)(2\sqrt{x}+7)=0`
`⇔ x=4\ (TM)`
Vậy `x=4` thì `3P=2\sqrt{x}+4`
6)
`P<4`
`⇔ \frac{3(\sqrt{x}+6)}{\sqrt{x}+1}<4`
`⇔ \frac{3(\sqrt{x}+6)}{\sqrt{x}+1}-4<0`
`⇔ \frac{3(\sqrt{x}+6)}{\sqrt{x}+1}-\frac{4\sqrt{x}+4}{\sqrt{x}+1}<0`
`⇔ \frac{3\sqrt{x}+18-4\sqrt{x}-4}{\sqrt{x}+1}<0`
`⇔ \frac{14-\sqrt{x}}{\sqrt{x}+1}<0`
Ta có: `x \ge 0⇒\sqrt{x} \ge 0⇒\sqrt{x}+1 \ge 1 \forall x`
`⇔ 14-\sqrt{x} < 0`
`⇔ 14<\sqrt{x}`
`⇔ 196<x` kết hợp ĐK
Vậy `x > 196` thì `P<4`
