\(y+2⋮x;x+2⋮y\Rightarrow\left(x+2\right)\left(y+2\right)⋮xy\Rightarrow xy+2x+2y+4⋮xy\Rightarrow2x+2y+4⋮xy\)

\(\Rightarrow2\left(x+y+2\right)⋮xy\Rightarrow2⋮xy\Rightarrow xy\inƯ\left(2\right)=1;2\)

\(xy=1\Rightarrow x=1,y=1\Rightarrow y+2=1+2=3⋮x=1\Rightarrow y+2⋮x\)

                                             \(x+2=1+2=3⋮y=1\Rightarrow x+2⋮y\)

\(\Rightarrow x=1,y=1\left(tm\right)\)

\(xy=2\Rightarrow x=1,y=2;x=2,y=1\Rightarrow x+2=1+2=3\)ko chia hết cho \(y=2\Rightarrow x+2\)ko chia hết cho y

\(\Rightarrow x=1,y=2\left(ktm\right)\Rightarrow x=2,y=1\left(ktm\right)\)

vậy x=1,y=1 

ko chắc lắm

Câu 2: 

a: (x+3)(y+2)=1

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

hay \(\left(x,y\right)\in\left\{\left(-4;-3\right);\left(-2;-3\right)\right\}\)

b: (2x-5)(y-6)=17

\(\Leftrightarrow\left(2x-5;y-6\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(3;23\right);\left(11;7\right);\left(2;-11\right);\left(-6;5\right)\right\}\)

c: \(\left(x-1\right)\left(x+y\right)=33\)

\(\Leftrightarrow\left(x-1;x+y\right)\in\left\{\left(1;33\right);\left(33;1\right);\left(-1;-33\right);\left(-33;-1\right);\left(3;11\right);\left(11;3\right);\left(-11;-3\right);\left(-3;-11\right)\right\}\)

hay \(\Leftrightarrow\left(x;x+y\right)\in\left\{\left(2;33\right);\left(34;1\right);\left(0;-33\right);\left(-32;-1\right);\left(4;11\right);\left(12;3\right);\left(-10;-3\right);\left(-2;-11\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(2;31\right);\left(34;-33\right);\left(0;-33\right);\left(-32;31\right);\left(4;7\right);\left(12;-9\right);\left(-10;7\right);\left(-2;-9\right)\right\}\)

ai TL đc giúp tui đi

a. x + xy + y = 9

=> xy + x + y - 9 = 0

=> xy + x + y + 1 - 10 = 0

=> x.(y + 1) + (y + 1) = 10

=> (y + 1).(x + 1) = 10

Lập bảng:

Vậy các cặp (x;y) thỏa là: (-11;-2); (-6;-3); (-3;-6); (-2;-11); (0;9); (1;4); (4;1); (9;0).

b. 3x + 3 chia hết cho 2x + 1

=> 2.(3x + 3) chia hết cho 2x + 1

=> 6x + 6 chia hết cho 2x + 1

=> (6x + 6 - 2x - 1) chia hết cho 2x + 1

=> 4x + 6 chia hết cho 2x + 1

=> 4x + 2 + 4 chia hết cho 2x + 1

=> 2.(2x + 1) + 4 chia hết cho 2x + 1

Mà 2.(2x + 1) chia hết cho 2x + 1

=> 4 chia hết cho 2x + 1

=> 2x + 1 thuộc Ư(4) = {-4; -2; -1; 1; 2; 4}

=> x thuộc {-5/2; -3/2; -1; 0; 1/2; 3/2}

Mà x thuộc Z

=> x thuộc {-1; 0}.

Sửa lại câu b:

.....

=> (6x + 6 - 2x - 1) chia hết cho 2x + 1

=> 4x + 5 chia hết cho 2x + 1

=> 4x + 2 + 3 chia hết cho 2x + 1

=> 2.(2x + 1) + 3 chia hết cho 2x + 1

=> 3 chia hết cho 2x + 1

=> 2x + 1 thuộc Ư(3) = {-3; -1; 1; 3}

=> x thuộc {-2; -1; 0; 1}.

\(a,x-5⋮x+2\)

\(\Rightarrow x+2-7⋮x+2\)

\(\Rightarrow x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x + 2 = 1=> x = -1

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

.... tương tự nhé ~ 

\(2x+3⋮x-5\)

\(\Rightarrow2x-10+7⋮x-5\)

\(\Rightarrow2\left(x-5\right)+7⋮x-5\)

\(\Rightarrow x-5\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

x - 5 = 1 => x = 6 

.... 

\(xy+8=x+y\)

\(\Leftrightarrow xy-x-y+1=-7\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)=-7=1.\left(-7\right)=\left(-1\right).7\)

\(=\left(-7\right).1=7.\left(-1\right)\)

\(5x+7⋮x-2\)

\(\Rightarrow5\left(x-2\right)+17⋮x-2\)

\(\Rightarrow17⋮x-2\)

\(\Leftrightarrow x-2\in\left\{1;-1;17;-17\right\}\)

\(\Leftrightarrow x\in\left\{3;1;19\right\}\)