Question

tìm x,y nguyên sao cho

xy-2y+3(x-2)=7

Answer 1

\(xy-2y+3\left(x-2\right)=7\)

\(xy-2y+3x-3\cdot2=7\)

\(\left(x-2\right)y+\left(x-2\right)3=7\)

\(\left(x-2\right)\left(y+3\right)=7\)

Ta có: 1 x 7 = 7;              7 x1 = 7

          (-1) x (-7) = 7 ;        (-7) x (-1)= 7

=>  TH1:  x-2= 1 =>   x=2

              y+3 = 7 => y= 4

TH2:  x-2= 7 =>   x=9

              y+3 = 1 => y= -2

TH3:  x-2= -1 =>   x=1

              y+3 = -7 => y= -10

TH4:  x-2= -7 =>   x=-5

              y+3 = -1 => y= -4

Vậy \(x\in\left\{-5;1;2;9\right\}\)

\(y\in\left\{-10;-4;-2;4\right\}\)