Lời giải:

a) ĐKXĐ: \(\left\{\begin{matrix} x+1\neq 0\\ x-1\neq 0\\ 2-2x^2\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 1\)

b) 

\(A=\left[\frac{x(x-1)}{(x-1)(x+1)}+\frac{x+1}{(x+1)(x-1)}+\frac{2x}{(x-1)(x+1)}\right].\frac{1}{x+1}=\frac{x^2+2x+1}{(x-1)(x+1)}.\frac{1}{x+1}\)

\(=\frac{(x+1)^2}{(x-1)(x+1)}.\frac{1}{x+1}=\frac{1}{x-1}\)

Để $A$ nguyên thì $1\vdots x-1$

$\Rightarrow x-1\in\left\{\pm 1\right\}$

$\Rightarrow x\in\left\{0;2\right\}$ (đều thỏa mãn đkxđ)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

\(=\left(\dfrac{2x\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)\left(x-1\right)}+\dfrac{4x}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{1}{x+1}\)

\(=\dfrac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{1}{x+1}\)

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

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

\(=\dfrac{1}{x-1}\)

b) Để A nguyên thì \(1⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(1\right)\)

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

hay \(x\in\left\{2;0\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;0\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{2;0\right\}\)

\(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne-\frac{1}{2}\end{cases}}\)

a) \(A=\left(\frac{1}{x-1}+\frac{x}{x^3-1}\cdot\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)

\(\Leftrightarrow A=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)

\(\Leftrightarrow A=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)

\(\Leftrightarrow A=\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow A=\frac{x+1}{x-1}\)

b) Thay \(x=\frac{1}{2}\)vào A, ta được :

\(A=\frac{\frac{1}{2}+1}{\frac{1}{2}-1}=\frac{\frac{3}{2}}{-\frac{1}{2}}=-3\)

`a)ĐK:` \(\begin{cases}x \ge 0\\x-\sqrt{x} \ne 0\\x-1 \ne 0\\\end{cases}\)

`<=>` \(\begin{cases}x \ge 0\\x \ne 0\\x \ne 1\\\end{cases}\)

`<=>` \(\begin{cases}x>0\\x \ne 1\\\end{cases}\)

`b)A=(sqrtx/(sqrtx-1)-1/(x-sqrtx)):(1/(1+sqrtx)+2/(x-1))`

`=((x-1)/(x-sqrtx)):((sqrtx-1+2)/(x-1))`

`=(x-1)/(x-sqrtx):(sqrtx+1)/(x-1)`

`=(sqrtx+1)/sqrtx:1/(sqrtx-1)`

`=(x-1)/sqrtx`

`c)A>0`

Mà `sqrtx>0AAx>0`

`<=>x-1>0<=>x>1`

a, ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

b, Ta có : \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

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

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

c, Ta có : \(A>0\)

\(\Leftrightarrow x-1>0\)

\(\Leftrightarrow x>1\)

Vậy ...

a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3

a) Ta có: \(A=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\)

\(=\left(\dfrac{3\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right):\dfrac{2x-1}{4x-8}\)

\(=\dfrac{3x-6-2x^2-4x+4x^2+6}{2\left(x+2\right)\left(x-2\right)}\cdot\dfrac{4\left(x-2\right)}{2x-1}\)

\(=\dfrac{2x^2-x}{x+2}\cdot\dfrac{2}{2x-1}\)

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

\(=\dfrac{2x}{x+2}\)

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

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

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

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

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a) Ta có: \(A=\left(1+\dfrac{x^2}{x^2+1}\right):\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3+x-x^2-1}\right)\)

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

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

\(=\dfrac{2x^2+1}{x-1}\)

b) Thay \(x=-\dfrac{1}{2}\) vào A, ta được:

\(A=\left(2\cdot\dfrac{1}{4}+1\right):\left(\dfrac{-1}{2}-1\right)\)

\(=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)

c) Để A<1 thì A-1<0

\(\Leftrightarrow\dfrac{2x^2+1}{x-1}-1< 0\)

\(\Leftrightarrow\dfrac{2x^2+1-x+1}{x-1}< 0\)

\(\Leftrightarrow\dfrac{2x^2-x+2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\)

hay x<1

ĐKXĐ: \(x\notin\left\{2;-2;-1\right\}\)

a) Ta có: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{4}{2-x}+\dfrac{1}{x+2}\right):\dfrac{3x+3}{x^2+2x}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right):\dfrac{3\left(x+1\right)}{x\left(x+2\right)}\)

\(=\left(\dfrac{x+4x+8}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{5x+8+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6x+6}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

\(=\dfrac{6\left(x+1\right)}{x-2}\cdot\dfrac{x}{3\left(x+1\right)}\)

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

b) Để A nguyên thì \(2x⋮x-2\)

\(\Leftrightarrow2x-4+4⋮x-2\)

mà \(2x-4⋮x-2\)

nên \(4⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(4\right)\)

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

\(\Leftrightarrow x\in\left\{3;1;4;0;6;-2\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{0;1;3;4;6\right\}\)

Vậy: Khi \(x\in\left\{0;1;3;4;6\right\}\) thì A nguyên