Lời giải:
Cần bổ sung điều kiện $x$ là số nguyên.

a.

$2x+5\vdots x+1$

$\Rightarrow 2(x+1)+3\vdots x+1$

$\Rightarrow 3\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3\right\}$

$\Rightarrow x\in\left\{0; -2; 2; -4\right\}$

b.

$-x-5\vdots -x-1$

$\Rightarrow (-x-1)-4\vdots -x-1$

$\Rightarrow 4\vdots -x-1$

$\Rightarrow -x-1\in\left\{\pm 1; \pm 2; \pm 4\right\}$

$\Rightarrow x\in \left\{0; -2; 1; -3; 3; -5\right\}$

a: =>2x+2+3 chia hêt cho x+1

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

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

b: =>x+5 chia hết cho x+1

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

=>\(x\in\left\{0;-2;1;-3;3;-5\right\}\)

Bài 1: 

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

hay \(x\in\left\{2;0;4;-2\right\}\)

a: 3x^3+2x^2-7x+a chia hêt cho 3x-1

=>3x^3-x^2+3x^2-x-6x+2+a-2 chia hết cho 3x-1

=>a-2=0

=>a=2

c: =>2x^2-6x+(a+6)x-3a-18+3a+19 chia x-3 dư 4

=>3a+19=4

=>3a=-15

=>a=-5

d: 2x^3-x^2+ax+b chiahêt cho x^2-1

=>2x^3-2x-x^2+1+(a+2)x+b-1 chia hết cho x^2-1

=>a+2=0 và b-1=0

=>a=-2 và b=1

a: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

hay \(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

a: \(x\in\left\{1;7\right\}\)

b: \(x+1=1\)

hay x=0

\(a,\Rightarrow x\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ b,\Rightarrow2\left(x+1\right)-1⋮x+1\\ \Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{-2;0\right\}\)

a: 

b: 

hay x=0

Bài 1:

Ta có: \(5x^3-3x^2+2x+a⋮x+1\)

\(\Leftrightarrow5x^3+5x^2-8x^2-8x+10x+10+a-10⋮x+1\)

\(\Leftrightarrow a-10=0\)

hay a=10

\(\Leftrightarrow2x^3+6x^2-x^2-3x+6x+18+m-13⋮x+3\)

hay m=13