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}

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}

\(\frac{4x^2-6x+5}{2x-1}=2x-2+\frac{3}{2x-1}\)

Để biểu thức có giá trị nguyên thì \(\left(2x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

Với 2x - 1 = 1 => 2x = 2 => x = 1

      2x - 1 = -1 => 2x = 0 => x = 0

      2x - 1 = 3 => 2x = 4 => x = 2

      2x - 1 = -3 => 2x = -2 => x = -1

Vậy x = {1;0;2;-1}

Đặt \(A=\frac{3x^2-4x-17}{x+2}\)

\(A=\frac{3x^2+6x-10x-20+3}{x+2}=\frac{3x.\left(x+2\right)-10.\left(x+2\right)+3}{x+2}=\frac{\left(x+2\right).\left(3x-10\right)+3}{x+2}\)

                                                                   \(A=\frac{\left(x+2\right).\left(x-10\right)}{x+2}+\frac{3}{x+2}=x-10+\frac{3}{x+2}\)

Do x nguyên => x - 10 nguyên

Để A nguyên thì \(\frac{3}{x+2}\) nguyên

\(\Rightarrow3⋮x+2\)

\(\Rightarrow x+2\in\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{-1;-3;1;-5\right\}\)

Vậy \(x\in\left\{-1;-3;1;-5\right\}\) thì \(\frac{3x^2-4x-17}{x+2}\) nguyên