Ta có: \(xy+x-2y+3=x\left(y+1\right)-2y-2+5\)

                                     \(=x\left(y+1\right)-2\left(y+1\right)+5\)

                                  \(=\left(y+1\right)\left(x-2\right)+5\)

   (y+1)(x+2)+5=0

nên (y+1)(x-2)=(-5)= (-1) * 5 =5 * (-1)= 1 * (-5) = (-1)*5

 + Nếu y+1 = -1  ; x-2= 5 -> y= -2   ;  x =7

 + Nếu y+1 = 5 ; x-2=-1  -> y=4    ; x = 1

 + Nếu y+1 = 1 ; x-2=-5      ->    y=0    ; x=-3

+ Nếu y+1= -5 ; x-2=1   -> y=-6           ; x=3  

thank you nha mìn cũng đang tìm câu này

x-3 =3

   đáp số ; 3

ko bt làm :Đ

dấu ^ là j vậy bạn

\(x^2+xy-3x-2y+1=0\)

\(\Rightarrow x^2-2x+xy-2y-x+1=0\)

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

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

\(\Rightarrow x+y=\frac{x-1}{x-2}\)

\(Do\) \(x-1\)ko chia hết cho \(x-2\)

\(\Rightarrow\frac{x-1}{x-2}\notin Z\)

\(Hay\) \(x+y\notin Z\)

Vậy ko có giá trị x,y thỏa mãn đề bài

Mik cug k bt có đúng k 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)$

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