Mình gửi bạn nha

Đáp án+ Giải thích các bước giải: 

`a,` 

`P = ( {x\sqrt{x} -1}/{x -\sqrt{x}} - {x\sqrt{x} +1}/{x +\sqrt{x}} ):] {2(x -2\sqrt{x} +1)}/{x-1} ]`
`ĐK: x \ge 0; x \ne 1`

`P =[ {(\sqrt{x} -1)(x +\sqrt{x} +1)}/{\sqrt{x}(\sqrt{x} -1)} - {(\sqrt{x} +1)(x -\sqrt{x} +1)}/{\sqrt{x}(\sqrt{x} +1)} ] : {2(\sqrt{x} -1)^2}/{(\sqrt{x} -1)(\sqrt{x} +1)}`
`P = ( {x +\sqrt{x} +1}/{\sqrt{x}} - {x -\sqrt{x} +1}/{\sqrt{x}} ) : {2(\sqrt{x} -1}/{\sqrt{x} -1}`
`P = {x +\sqrt{x} +1 - x +\sqrt{x} -1}/{\sqrt{x}} . {\sqrt{x} +1}/{2(\sqrt{x} -1)}`
`P = {2\sqrt{x}}/{\sqrt{x}} . {\sqrt{x} +1}/{2(\sqrt{x} -1)}`
`P = {\sqrt{x} +1}/{\sqrt{x} -1}`
`b, P < 0`

`<=> {\sqrt{x} +1}/{\sqrt{x} -1} < 0`
Ta có: `\sqrt{x} \ge 0 => \sqrt{x} + 1 > 0`

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

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

`KHĐK: 0 \le x < 1`

Vậy: `0 \le x < 1` thì `P < 0`

`c,`

`P = {\sqrt{x} +1}/{\sqrt{x} -1}`
`P = {\sqrt{x} -1+2}/{\sqrt{x} -1}`
`P = {\sqrt{x} -1}/{\sqrt{x} -1} + 2/{\sqrt{x} -1}`
`P = 1 + 2/{\sqrt{x} -1}`
`P` nguyên `<=> 1 + 2/{\sqrt{x} -1} \in ZZ`
                      `=> 2/{\sqrt{x} -1} \in ZZ `

                      `=> \sqrt{x} - 1 \in Ư(2)`

                      `=> \sqrt{x} - 1 \in { +- 1; +-2}`
\begin{array}{|c|c|c|}\hline \text{$\sqrt{x} -1$}&\text{$-1$}&\text{1}&\text{$-2$}&\text{2}\\\hline \text{$\sqrt{x}$}&\text{0}&\text{2}&\text{$-1(loại)$}&\text{3}\\\hline \text{$x$}&\text{0(tm)}&\text{4(tm)}&\text{}&\text{9(tm)}\\\hline\end{array}

`-> x \in { 0;4;9}` thì $P$ nguyên.