\(1;A=\frac{x+7}{x+1}=\frac{x+1+6}{x+1}=1+\frac{6}{x+1}\)

Vậy x + 1 là ước của 6 \(\Rightarrow x+1\in\left(1;-1;2;-2;3;-3;6;-6\right)\)

\(\Rightarrow x\in\left(0;-2;1;-3;2;-4;5;-7\right)\)

\(2;A=\frac{6x-2}{2x-3}=\frac{6x-9+7}{2x-3}=3+\frac{7}{2x-3}\)

Vậy 2x - 3 là ước của 7 \(\Rightarrow2x-3\in\left(1;-1;7;-7\right)\)

\(\Rightarrow x\in\left(2;1;5;-2\right)\)

\(3;A=\frac{4x-8}{2x+1}=\frac{4x+2-10}{2x+1}=2-\frac{10}{2x+1}\)

Vậy 2x + 1 là ước của 10 => .........

a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)

+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)

\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)

+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)

+) \(x+1\ne0\Leftrightarrow x\ne-1\)

+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)

\(\Leftrightarrow x\ne0;x\ne-2\)

+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)

Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)

a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)

\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)

\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)

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

b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)

Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vậy ............................

b) \(A=\left(\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right):\frac{x^2-4}{3x^2+6x}\)

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

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

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

\(=\frac{x^3+1}{x^3+1}\)\(.\frac{3x^2+6x}{x^2-4}\)\(=\frac{3x^2+6x}{x^2-4}\)

\(=\frac{3x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{3x}{x-2}\)

A nguyên\(\Leftrightarrow3x⋮x-2\)

\(\Leftrightarrow3\left(x-2\right)+6⋮x-2\)

Mà \(\left(x-2\right)⋮x-2\Rightarrow6⋮x-2\)

\(\Rightarrow x-2\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Lập bảng:

Vậy\(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)