xy + 2y - x -y = 5

<=> xy + y - x = 5

<=> xy + y - x -1 = 5 -1

<=> (xy + y) - (x +1)= 4

<=> y (x+1) - (x+1) = 4

<=> (x +1) (y -1) = 4

ta có 4 = 4.1 hoặc 4 = 2.2

Vậy các cặp x, y thỏa mãn là:

x = 0; y= 5

x = 3; y = 2

x = 1; y = 3

=> (y+1)x + 2y = 5

=> (y+1)x+2y+2=7

=>(y+1)x+2(y+1)=7

=>(y+1)(x+2) = 7

Do, x,y thuộc N nên ta xét:

TH1: y+1=1, x+2=7=> y=0, x=5

TH2: y+1=7, x+2=1=> x=6,x=-1(loại)

vậy y=0 và x=5

Ta có : 

\(xy+x+2y=5\)

\(\Rightarrow\left(xy+2y\right)+x+2=7\)

\(\Rightarrow y\left(x+2\right)+\left(x+2\right)=7\)

\(\Rightarrow\left(x+2\right)\left(y+1\right)=7\)

Do \(x;y\in N\)

\(\Rightarrow x+2;y+1\in N\)

Mà \(x+2;y+1\inƯ\left(7\right)\)

\(\Rightarrow x+2;y+1\in\left\{1;7\right\}\)

Ta có bảng sau : 

Vậy \(x=5;y=0\)

a) \(xy+x+2y=5\\ \Rightarrow y\left(x+2\right)+x+2=5+2\\ \Rightarrow\left(x+2\right)\left(y+1\right)=7\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;6\right);\left(5;0\right);\left(-3;-8\right);\left(-9;-2\right)\right\}\)

b) \(xy-3x-y=0\\ \Rightarrow x\left(y-3\right)-y+3=3\\ \Rightarrow\left(y-3\right)\left(x-1\right)=3\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(2;6\right);\left(4;4\right);\left(0;0\right);\left(-2;2\right)\right\}\)

c) \(xy+2x+2y=-16\\ \Rightarrow x\left(y+2\right)+2y+4=-12\\ \Rightarrow\left(y+2\right)\left(x+2\right)=-12\)

Ta xét bảng:

Vậy \(\left(x;y\right)\in\left\{\left(-1;-14\right);\left(0;-8\right);\left(1;-6\right);\left(2;-5\right);\left(4;-4\right);\left(10;-3\right);\left(-3;10\right);\left(-4;4\right);\left(-5;2\right);\left(-6;1\right);\left(-8;0\right);\left(-14;-1\right)\right\}\)