\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)

\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)

+ Với \(x+\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)

+ Với \(x+\sqrt{x}+1=2\)

\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)

a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1

A=(\(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

A=\(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)

A=\(\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)=\(\dfrac{2}{a-1}\)

b) Để A\(\in\)Z\(\Rightarrow\)x-1\(\in\) Ư(2)=\(\left\{-1,1,-2,2\right\}\)

vì x\(\ge\)0,x\(\ne\)1 nên x\(\in\)\(\left\{-1,0,2,3\right\}\)

Lời giải:

ĐK: $a>0; a\neq 4$

\(A=\frac{(\sqrt{a}+2)(\sqrt{a}-2)}{a}-1=\frac{a-4}{a}-1=\frac{-4}{a}\)

Với $a$ nguyên, để $A$ nhận giá trị nguyên thì $-4\vdots a$

Mà $a>0; a\neq 4$ nên $a=1$ hoặc $a=2$

a: ĐKXĐ: x=0; x<>1

\(M=\left(2+\sqrt{x}\right)\left(1-2\sqrt{x}-x+1+\sqrt{x}+x\right)\)

\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)=4-x\)

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

Để A là số nguyên dương thì \(\left\{{}\begin{matrix}3\sqrt{x}+6-7⋮\sqrt{x}+2\\x>\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\sqrt{x}+2=7\)

hay x=25

Bài 8:

\(M=1+\frac{4}{\sqrt{x}+1}\)

Để $M$ nguyên thì $\frac{4}{\sqrt{x}+1}$ nguyên 

Đặt $\frac{4}{\sqrt{x}+1}=t$ với $t$ là số nguyên dương 

$\Rightarrow \sqrt{x}+1=\frac{4}{t}$

$\sqrt{x}=\frac{4}{t}-1=\frac{4-t}{t}\geq 0$

$\Rightarrow 4-t\geq 0\Rightarrow t\leq 4$

Mà $t$ nguyên dương suy ra $t=1;2;3;4$

Kéo theo $x=9; 1; \frac{1}{9}; 0$

Kết hợp đkxđ nên $x=0; \frac{1}{9};9$

Bài 9:

$P=1+\frac{5}{\sqrt{x}+2}$

Để $P$ nguyên thì $\frac{5}{\sqrt{x}+2}$ nguyên 

Đặt $\frac{5}{\sqrt{x}+2}=t$ với $t\in\mathbb{Z}^+$

$\Leftrightarrow \sqrt{x}+2=\frac{5}{t}$

$\Leftrightarrow \sqrt{x}=\frac{5-2t}{t}\geq 0$

Với $t>0\Rightarrow 5-2t\geq 0$

$\Leftrightarrow t\leq \frac{5}{2}$

Vì $t$ nguyên dương suy ra $t=1;2$

$\Rightarrow x=9; \frac{1}{4}$ (thỏa đkxđ)

Bài 8: 

Để M nguyên thì \(\sqrt{x}+5⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3\right\}\)

hay \(x\in\left\{0;1;9\right\}\)

(a) Với \(x\ge0,x\ne4\), ta có: 

\(A=\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)

Để \(A\le5\Rightarrow2\sqrt{x}+1\le5\)

\(\Leftrightarrow2\sqrt{x}\le4\Leftrightarrow\sqrt{x}\le2\Leftrightarrow0\le x\le4\).

Kết hợp với điều kiện thì: \(0\le x< 4.\)

(b) \(\dfrac{A}{2}=\dfrac{2\sqrt{x}+1}{2}\) nguyên khi \(\left(2\sqrt{x}+1\right)\in B\left(2\right)=\left\{0;2;4;...;2n\right\}\left(n\in N\right)\)

\(\Leftrightarrow\sqrt{x}\in\left\{-\dfrac{1}{2};\dfrac{1}{2};\dfrac{3}{2};...;\dfrac{2n+1}{2}\right\}\left(n\in N\right)\)

Hay: \(\sqrt{x}\in\left\{\dfrac{1}{2};\dfrac{3}{2};...;\dfrac{2n+1}{2}\right\}\)

\(\Leftrightarrow x\in\left\{\dfrac{1}{4};\dfrac{9}{4};...;\dfrac{\left(2n+1\right)^2}{4}\right\}\)

Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để A nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)

\(\Leftrightarrow2⋮\sqrt{x}-2\)

\(\Leftrightarrow\sqrt{x}-2\in\left\{-2;-1;1;2\right\}\)

Vậy: Có 4 giá trị nguyên của x thỏa mãn yêu cầu đề bài

a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)