1.     Giải:

Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)

 \(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)

 \(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)

Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.

⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)

Ta có bảng:

Vậy xϵ\(\left\{0;1;3;10\right\}.\)

2. Giải:

Do (2x-18).(3x+12)=0.

⇒ 2x-18=0             hoặc             3x+12=0.

⇒ 2x     =18                               3x       =-12.

⇒   x     =9                                   x       =-4.

Vậy xϵ\(\left\{-4;9\right\}.\)

3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.

S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.

S= 0 + 0 + ... + 0 + 2025.

⇒S= 2025.

a) \(x^2+x+1=x\left(x+1\right)+1\)

Vì \(x\inℤ\)\(\Rightarrow x\left(x+1\right)⋮x+1\)\(\Rightarrow\)Để \(x^2+x+1⋮x+1\)thì \(1⋮x+1\)

\(\Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\)\(\Rightarrow x\in\left\{-2;0\right\}\)

Vậy \(x\in\left\{-2;0\right\}\)

b) \(3x-8=3x-12+4=3\left(x-4\right)+4\)

Vì \(3\left(x-4\right)⋮x-4\)\(\Rightarrow\)Để \(3x-8⋮x-4\)thì \(4⋮x-4\)

\(\Rightarrow x-4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Lập bảng giá trị ta có: 

Vậy \(x\in\left\{0;2;3;5;6;8\right\}\)

Bài 10:

a: 2x-3 là bội của x+1

=>\(2x-3⋮x+1\)

=>\(2x+2-5⋮x+1\)

=>\(-5⋮x+1\)

=>\(x+1\in\left\{1;-1;5;-5\right\}\)

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

b: x-2 là ước của 3x-2

=>\(3x-2⋮x-2\)

=>\(3x-6+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\inƯ\left(4\right)\)

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

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

Bài 14:

a: \(4n-5⋮2n-1\)

=>\(4n-2-3⋮2n-1\)

=>\(-3⋮2n-1\)

=>\(2n-1\inƯ\left(-3\right)\)

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

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

=>\(n\in\left\{1;0;2;-1\right\}\)

mà n>=0

nên \(n\in\left\{1;0;2\right\}\)

b: \(n^2+3n+1⋮n+1\)

=>\(n^2+n+2n+2-1⋮n+1\)

=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)

=>\(-1⋮n+1\)

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

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

mà n là số tự nhiên

nên n=0

Bài 16:

a: \(\left(x+5\right)\left(y-3\right)=15\)

=>\(\left(x+5\right)\left(y-3\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>\(\left(x+5;y-3\right)\in\left\{\left(1;15\right);\left(15;1\right);\left(-1;-15\right);\left(-15;-1\right);\left(3;5\right);\left(5;3\right);\left(-3;-5\right);\left(-5;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-4;18\right);\left(10;4\right);\left(-6;-12\right);\left(-20;2\right);\left(-2;8\right);\left(0;6\right);\left(-8;-2\right);\left(-10;0\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(10;4\right);\left(0;6\right)\right\}\)

b: x là số tự nhiên

=>2x-1 lẻ và 2x-1>=-1

\(\left(2x-1\right)\left(y+2\right)=24\)

mà 2x-1>=-1 và 2x-1 lẻ

nên \(\left(2x-1\right)\cdot\left(y+2\right)=\left(-1\right)\cdot\left(-24\right)=1\cdot24=3\cdot8\)

=>\(\left(2x-1;y+2\right)\in\left\{\left(-1;-24\right);\left(1;24\right);\left(3;8\right)\right\}\)

=>\(\left(2x;y\right)\in\left\{\left(0;-26\right);\left(2;22\right);\left(4;6\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;-26\right);\left(1;11\right);\left(2;6\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(1;11\right);\left(2;6\right)\right\}\)

c:

x,y là các số tự nhiên

=>x+3>=3 và y+2>=2

xy+2x+3y=0

=>\(xy+2x+3y+6=6\)

=>\(x\left(y+2\right)+3\left(y+2\right)=6\)

=>\(\left(x+3\right)\left(y+2\right)=6\)

mà x+3>=3 và y+2>=2

nên \(\left(x+3\right)\cdot\left(y+2\right)=3\cdot2\)

=>x=0 và y=0

d: xy+x+y=30

=>\(xy+x+y+1=31\)

=>\(x\left(y+1\right)+\left(y+1\right)=31\)

=>\(\left(x+1\right)\left(y+1\right)=31\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(y+1\right)=1\cdot31=31\cdot1=\left(-1\right)\cdot\left(-31\right)=\left(-31\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y+1\right)\in\left\{\left(1;31\right);\left(31;1\right);\left(-1;-31\right);\left(-31;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right);\left(-2;-32\right);\left(-32;-2\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right)\right\}\)

a) x+3=(x-4)+4+3

         = (x-4)+7

Ta có:

x+3 chia hét cho x-4

=> (x-4)+7 chia het cho x-4

Mà x-4 chia het cho x-4

=> 7 chia het cho x-4

=> x-4 thuộc ước của 7={1;-1;7;-7}

tiếp theo rồi bn tự kẻ bảng làm nhé

\(\Rightarrow2\left(x+1\right)-5⋮x+1\\ \Rightarrow x+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow x\in\left\{-6;-2;0;4\right\}\)

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

hay \(x\in\left\{0;-2;4;-6\right\}\)

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

hay \(x\in\left\{0;-2;4;-6\right\}\)

\(\Rightarrow2\left(x+1\right)-5⋮\left(x+1\right)\\ \Rightarrow5⋮\left(x+1\right)\\ \Rightarrow x+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow x\in\left\{-6;-2;0;4\right\}\)

2x-3 là bội của x+1 

\(\Rightarrow2x-3⋮x+1\\ \Rightarrow2\left(x+1\right)-5⋮x+1\)

mà \(2\left(x+1\right)⋮x+1\forall x\\ \)

\(\Rightarrow5⋮x+1\\ \Rightarrow x+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=5\\x+1=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=4\\x=-6\end{matrix}\right.\)