a) Rút gọn b) Cm: P<1/3 P= $\frac{x+2}{x\sqrt[]{x}-1}$ + $\frac{\sqrt[]{x}+1}{x+\sqrt[]{x}+1}$ -$\frac{\sqrt[]{x}+1}{x-1}$

Đáp án:

Giải thích các bước giải:

`P=(x+2)/(x\sqrt{x}-1)+(\sqrt{x}+1)/(x+\sqrt{x}+1)-(\sqrt{x}+1)/(x-1)(ĐKXĐ:x>=0;xne1)`

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

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

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

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

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

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

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

`b,P<1/3`

`<=>P-1/3<0`

`<=>\sqrt{x}/(x+\sqrt{x}+1)-1/3<0`

`<=>3\sqrt{x}-(x+\sqrt{x}+1)<0`

`<=>3\sqrt{x}-x-\sqrt{x}-1<0`

`<=>-x+2\sqrt{x}-1<0`

`<=>-(x-2\sqrt{x}+1)<0`

`<=>(\sqrt{x}-1)^2>0` (luôn đúng do `xne1`)

Vậy `P<1/3`