Đáp án+Giải thích các bước giải:

4,

a,

`P=(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}):(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{1}{2})`

`Đk:`

$\begin{cases}x≥0\\x-9\neq0\\\sqrt{x}-3\neq0\end{cases}$

$⇔\begin{cases}x≥0\\x\neq9\\x\neq9\end{cases}$

Vậy `x≥0;x\ne9` thì biểu thức xác định

b,

`P=(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}):(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{1}{2})(x≥0;x\ne9)`

`P=(\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-3x-3}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\frac{2\sqrt{x}-2-\sqrt{x}+3}{2(\sqrt{x}-3)})`

`P=(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{(\sqrt{x}-3)(\sqrt{x}+3)}):\frac{\sqrt{x}+1}{2(\sqrt{x}-3)}`

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

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

`P=\frac{-6}{\sqrt{x}+3}`

c,

`P∈Z⇔-6\vdots\sqrt{x}+3⇔\sqrt{x}+3∈Ư(-6)={±1;±2;±3;±6}`

`\sqrt{x}+3≥3⇒\sqrt{x}+3∈{3,6}`

Lập bảng giá trị:
$\begin{array}{|c|c|c|}\hline \sqrt{x}+3&3&6\\\hline x&0&9(L)\\\hline \end{array}$

Vậy `x=0` thì `P∈Z`

Đáp án+Giải thích các bước giải:

`P=(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}):(\frac{\sqrt{x}-1}{\sqrt{x}-3}-1/2)`

ĐK: `{(\sqrt{x}-3\ne0),(x>=0):}`

`<=>{(x\ne9),(x>=0):}`

Vậy `ĐKXĐ: x\ne9;x>=0`

`b)P=(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}):(\frac{\sqrt{x}-1}{\sqrt{x}-3}-1/2)`

`=(\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-3x-3}{(\sqrt{x}-3)(\sqrt{x}+3)}):(\frac{2\sqrt{x}-2-\sqrt{x}+3}{2\sqrt{x}-6})`

`=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{(\sqrt{x}+3)(\sqrt{x}-3)}.\frac{2\sqrt{x}-6}{\sqrt{x}+1}`

`=\frac{(-3\sqrt{x}-3)(2\sqrt{x}-6)}{(\sqrt{x}+3)(\sqrt{x}-3)(\sqrt{x}+1)}`

`=\frac{-6(\sqrt{x}+1)(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)(\sqrt{x}+1)}`

`=\frac{-6}{\sqrt{x}+3}`

`c)P=\frac{-6}{\sqrt{x}+3}`

`=-\frac{6}{\sqrt{x}+3}`

Để `P\inZZ=>6\vdots\sqrt{x}+3`

`<=>\sqrt{x}+3\inƯ(6)\in{+-1;+-2;+-3;+-6}`

Mà `\sqrt{x}+3>=3`

`=>\sqrt{x}+3\in{3;6}`

`=>\sqrt{x}\in{3-3;6-3}`

`=>x\in{0;9}`

Mà `x\ne9`

`=>x=0` 

Vậy với `x=0` thì `P\inZZ`