Để N nguyên thì \(3x^2-4x-17⋮x+2\)

\(3x^2+6x-10x-20+3⋮x+2\)

\(3x\left(x+2\right)-10\left(x+2\right)+3⋮x+2\)

\(\left(x+2\right)\left(3x-10\right)+3⋮x+2\)

Dễ thấy \(\left(x+2\right)\left(3x-10\right)⋮x+2\)

\(\Rightarrow3⋮x+2\)

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

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

Vậy......

a: Để A là số nguyên thì \(x^3-3x^2-x^2+3x+x-3-7⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{4;2;10;-4\right\}\)

b: Đề sai rồi bạn

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

a)Với  x \(\ne\)-1

Ta có: x² + x = 0

=> x(x + 1) = 0

=> \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-1\left(ktm\right)\end{cases}}\)

Với x = 0 => A = \(\frac{0-3}{0+1}=-3\)

b) Ta có: B = \(\frac{3}{x-3}+\frac{6x}{9-x^3}+\frac{x}{x+3}\)

B = \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

B = \(\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

B = \(\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

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

B = \(\frac{x+3}{x-3}\)

c)  Với x \(\ne\)\(\pm\)3; x \(\ne\)-1

Ta có: P = AB = \(\frac{x-3}{x+1}\cdot\frac{x+3}{x-3}=\frac{x+3}{x+1}=\frac{\left(x+1\right)+2}{x+1}=1+\frac{2}{x+1}\)

Để P \(\in\)Z <=> 2 \(⋮\)x + 1

<=> x + 1 \(\in\)Ư(2) = {1; -1; 2; -2}

<=> x \(\in\){0; -2; 1; -3}