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: \(x\in\left\{2;10\right\}\)

ĐKXĐ: x<>1

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

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

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

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

=>\(\left(x-1\right)^2\in\left\lbrace1;3\right\rbrace\)

mà x ∈Z

nên \(\left(x-1\right)^2=1\)

=>x-1∈{1;-1}

=>x∈{2;0}

ĐKXĐ: x<>1

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

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

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

=>\(\left(x-1\right)^2=1\)

=>x-1∈{1;-1}

=>x∈{2;0}

c: Để \(\frac{x^2-x}{x-3}\) là số nguyên thì \(x^2-x\vdots x-3\)

=>\(x^2-3x+2x-6+6\vdots x-3\)

=>6⋮x-3

=>x-3∈{1;-1;2;-2;3;-3;6;-6}

=>x∈{4;2;5;1;6;0;9;-3}

d: \(\frac{3x^2-4x-15}{x+2}\) là số nguyên

=>\(3x^2-4x-15\vdots x+2\)

=>\(3x^2+6x-10x-20+5\vdots x+2\)

=>5⋮x+2

=>x+2∈{1;-1;5;-5}

=>x∈{-1;-3;3;-7}

e:

ĐKXĐ: x<>0; x<>-2; x<>2

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

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

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

Để A nguyên thì -2⋮x-2

=>x-2∈{1;-1;-2;2}

=>x∈{3;1;0;4}

Kết hợp ĐKXĐ, ta được: x∈{3;1;4}

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

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

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

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

ĐKXĐ: x<>3

\(\frac{4x^3-3x^2+2x-83}{x-3}\)

\(=\frac{4x^3-12x^2+9x^2-27x+29x-87+4}{x-3}=4x^2+9x+29+\frac{4}{x-3}\)

Để đây là số nguyên thì 4⋮x-3

=>x-3∈{1;-1;2;-2;4;-4}

=>x∈{4;2;5;1;7;-1}

a: Để A là số nguyên thì

x^3-2x^2+4 chia hết cho x-2

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

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

b: Để B là số nguyên thì

\(3x^3-x^2-6x^2+2x+9x-3+2⋮3x-1\)

=>\(3x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)