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 \(x\in\left\{2;0\right\}\)

a: ĐKXĐ: x<>2; x<>0

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

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

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

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

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7

1: Ta có: \(A=\left(\dfrac{x^2-16}{x-4}-1\right):\left(\dfrac{x-2}{x-3}+\dfrac{x+3}{x+1}+\dfrac{x+2-x^2}{x^2-2x-3}\right)\)

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

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

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

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

\(=x+1\)

ĐKXĐ: \(x\notin\left\{4;3;-1\right\}\)

2: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì \(x+1⋮x^2+x+1\)

\(\Leftrightarrow x^2+x⋮x^2+x+1\)

\(\Leftrightarrow x^2+x+1-1⋮x^2+x+1\)

mà \(x^2+x+1⋮x^2+x+1\)

nên \(-1⋮x^2+x+1\)

\(\Leftrightarrow x^2+x+1\inƯ\left(-1\right)\)

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

\(\Leftrightarrow x^2+x\in\left\{0;-2\right\}\)

\(\Leftrightarrow x^2+x=0\)(Vì \(x^2+x>-2\forall x\))

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

Vậy: Để \(\dfrac{A}{x^2+x+1}\) nhận giá trị nguyên thì x=0

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

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

b: A nguyên

=>x^2-1+3 chia hết cho x-1

=>\(x-1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;0;4;-2\right\}\)

a) đặt mẫu chứng là x-2

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

1. ĐKXĐ: \(x\ne0;x\ne2\)

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

\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)

\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}-\dfrac{2x^2}{\left(4+x^2\right)\left(2-x\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)

\(A=\left[\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(4+x^2\right)\left(x-2\right)}\right]\left(\dfrac{x^2-x-2}{x^2}\right)\)

\(A=\dfrac{x\left(x-2\right)^2+2.2x^2}{2\left(x^2+4\right)\left(x-2\right)}.\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{x^2}\)

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

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

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

\(A=\dfrac{x+1}{2x}\)