cho P= $\frac{3x+√9x-3}{x+√x-2}$ – $\frac{√x+1}{√x+2}$ + $\frac{√x-2}{1-√x}$ (x ≥0,x $\neq$ 1) a,rút gọn p b,so sánh p với √p với đk √p có nghĩa c,tì

`a)`

Ta có:`x+\sqrt{x}-2=x-\sqrt{x}+2\sqrt{x}-2`

`=\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)`

`(\sqrt{x}+2)(\sqrt{x}-1)`

`P={3x+\sqrt{9x}-3}/{x+\sqrt{x}-2}-{\sqrt{x}+1}/{\sqrt{x}+2}+{\sqrt{x}-2}/{1-\sqrt{x}}` $(x\ge 0; x\ne 1)$

`P={3x+3\sqrt{x}-3}/{(\sqrt{x}+2)(\sqrt{x}-1)}-{(\sqrt{x}+1)(\sqrt{x}-1)}/{(\sqrt{x}+2)(\sqrt{x}-1)}-{(\sqrt{x}-2)(\sqrt{x}+2)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`

`P={3x+3\sqrt{x}-3-(x-1)-(x-4)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`

`P={x+3\sqrt{x}+2}/{(\sqrt{x}+2)(\sqrt{x}-1)}`

`P={x+2\sqrt{x}+\sqrt{x}+2}/{(\sqrt{x}+2)(\sqrt{x}-2)}`

`P={(\sqrt{x}+1)(\sqrt{x}+2)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`

`P={\sqrt{x}+1}/{\sqrt{x}-1}`

`b)` `P={\sqrt{x}+1}/{\sqrt{x}-1}={(\sqrt{x}+1)(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`

`P={(\sqrt{x}+1)^2}/{x-1}`

`\sqrt{P}` có nghĩa khi $x-1>0\Leftrightarrow x>1$

+) `\sqrt{P}=\sqrt{{(\sqrt{x}+1)^2}/{x-1}}={\sqrt{x}+1}/ \sqrt{x-1}` $(x>1)$

+) `P-\sqrt{P}={\sqrt{x}+1}/{\sqrt{x}-1}-{\sqrt{x}+1}/ \sqrt{x-1}`

`=(\sqrt{x}+1)(1/{\sqrt{x}-1}-1/{\sqrt{x-1}})`

+) `x>1=>\sqrt{x}>1=>2\sqrt{x}>2=>-2\sqrt{x}<-2`

`=>x-2\sqrt{x}+1<x-2+1`

`=>(\sqrt{x}-1)^2<x-1=(\sqrt{x-1})^2`

`=>\sqrt{x}-1<\sqrt{x-1}`

`=>1/{\sqrt{x}-1}>1/{\sqrt{x-1}}`

`=>1/{\sqrt{x}-1}-1/{\sqrt{x-1}} >0`

`=>P-\sqrt{P}=(\sqrt{x}+1)(1/{\sqrt{x}-1}-1/{\sqrt{x-1}})>0`

`=>P>\sqrt{P}` với $x>1$

`c)`

`1/P={\sqrt{x}-1}/{\sqrt{x}+1}={\sqrt{x}+1-2}/{\sqrt{x}+1}=1-2/{\sqrt{x}+1`

Để `1/P \in Z=>(\sqrt{x}+1)\in Ư(2)={-2;-1;1;2}`

Mà `\sqrt{x}+1\ge 1 \forall x\ge 0`

`=>\sqrt{x}+1\in {1;2}`

`=>\sqrt{x}\in {0;1}`

`=>x\in {0;1}`

Kết hợp điều kiện `x\ne 1=>x=0`

Vậy $x=0$ thì `1/P \in Z`

Trả lời