a, \(x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(2x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

c, \(\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\Rightarrow x-1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

d, \(\dfrac{4\left(x-3\right)+3}{-\left(x-3\right)}=-4-\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

ĐKXĐ: \(x\ne1\)

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

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

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

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

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

Để B nguyên thì \(3⋮\left(x-1\right)^2\)

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

\(\Leftrightarrow\left(x-1\right)^2\in\left\{1;3;-1;-3\right\}\)

mà \(\left(x-1\right)^2>0\forall x\) thỏa mãn ĐKXĐ

nên \(\left(x-1\right)^2\in\left\{1;3\right\}\)

\(\Leftrightarrow x-1\in\left\{1;9\right\}\)

hay \(x\in\left\{2;10\right\}\) (nhận)

Vậy: \(x\in\left\{2;10\right\}\)

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

\(C\in Z\Leftrightarrow2⋮\left(x^2+2\right)\)

\(\Leftrightarrow x^2+2=2\Rightarrow x=0\)

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

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

|x+3|=5

=>x=2(loại) hoặc x=-8(nhận)

Khi x=-8 thì \(A=\dfrac{4}{64-8+2}=\dfrac{4}{58}=\dfrac{2}{29}\)

b: A nguyên

=>x^2+x+2 thuộc {1;-1;2;-2;4;-4}

=>x^2+x+2=2 hoặc x^2+x+2=4

=>x^2+x-2=0 hoặc x(x+1)=0

=>\(x\in\left\{1;0;-1\right\}\)