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

Để \(A\in Z\Leftrightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

<=>x thuộc {4;2;5;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

a: Để A là số nguyên thì \(x-3\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;2;5;1\right\}\)

b: Để B là số nguyên thì \(x+2-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

d: Để D là số nguyên thì \(3x^2+2x-3x-2+3⋮3x+2\)

\(\Leftrightarrow3x+2\in\left\{1;-1;3;-3\right\}\)

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

Ta có:\(x^2+4x+10=\left(x^2+2\cdot2\cdot x+2^2\right)+6=\left(x+2\right)^2+6\)

\(\Rightarrow\frac{3}{x^2+4x+10}=\frac{3}{\left(x+2\right)^2+6}\)

Do \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+6\ge6\)

\(\Rightarrow\frac{3}{\left(x+2\right)^2+6}\le\frac{3}{6}=\frac{1}{2}\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left(x+2\right)^2=0\Leftrightarrow x=-2\)

Vậy \(A_{min}=\frac{1}{2}\Leftrightarrow x=-2\)