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

thanks các bạn nhìu nha

a) \(xy+3x-2y-11=0\)

\(x\left(y+3\right)-2y-6-5=0\)

\(x\left(y+3\right)-2\left(y+3\right)=5\)

\(\left(x-2\right)\left(y+3\right)=5\)

\(x-2;y+3\in U\left(5\right)\)

b) \(xy+2x+y+11=0\)

\(x\left(y+2\right)+y+2+9=0\)

\(x\left(y+2\right)+\left(y+2\right)=-9\)

\(\left(x+1\right)\left(y+2\right)=-9\)

\(x+1;y+2\in U\left(-9\right)\)

a) $xy+3x-2y-11=0$$x\left(y+3\right)-2y-6-5=0$$x\left(y+3\right)-2\left(y+3\right)=5$$\left(x-2\right)\left(y+3\right)=5$$x-2;y+3\in U\left(5\right)$

b) $xy+2x+y+11=0$

$x\left(y+2\right)+y+2+9=0$$x\left(y+2\right)+\left(y+2\right)=-9$$\left(x+1\right)\left(y+2\right)=-9$$x+1;y+2\in U\left(-9\right)$

\(xy=2x+2y\\ \Rightarrow xy-2x-2y=0\\ \Rightarrow x\left(y-2\right)-2y+4=4\\ \Rightarrow x\left(y-2\right)-2\left(y-2\right)=4\\ \Rightarrow\left(x-2\right)\left(y-2\right)=4\)

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-2,y-2\in Z\\x-2,y-2\inƯ\left(4\right)\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(1;-2\right);\left(0;0\right);\left(-2;1\right);\left(3;6\right);\left(4;4\right);\left(6;3\right)\right\}\)

dẽ mà

xy-2(xy)=0

xy(2-1)=0

xy=0

x=0 hoặc y=0

a,Mình chỉ biết có 6 trường hợp của bài a thôi :)

TH1: x=4 ; y = -5

SORRY thiếu

TH2: x = -5 ; y = 4

TH3: x = 2 ; y = -10

TH4: x = -10 ; y = 2

TH5: x = 1 ; y = -20

TH6: x = -20 ; y = 1

TH1: x=4 ; y = -5

TH2: x = -5 ; y = 4

TH3: x = 2 ; y = -10

TH4: x = -10 ; y = 2

TH5: x = 1 ; y = -20

TH6: x = -20 ; y = 1

TH1: x=4 ; y = -5

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)

c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)

d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)

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