Question

Tìm hai số nguyên x và y : biết x+y=x.y

Answer 1

x + y = x . y

=> xy - x  - y = 0 

x ( y - 1 ) - ( y -1 ) = 0 +1

( x -1 ) ( y -1 )  = 1

ta có : 1 = 1 .1 = ( -1 ) . ( -1 )

T/H1 : x -1 = 1=> x = 2

       => y - 1 = 1 = > x =2

T/H2 : x -1 = -1 => x = 0

     => y -1 = -1 => y = 0

Vậy ( x ; y ) \(\in\){ ( 2 ; 2 ) ; ( 0 ; 0 }

Answer 2

x+y - x+y =0

[x - xy]+y-1=-1

x.[1-y]-[-y+1]=-1

x.[1-y]-[1-y]=-1

[1-y] .[x-1]=-1

ta thay y thuoc z suy ra 1-y thuoc z

ta thay x thuoc z suy ra x-1 thuoc z 

nen 1-y thuoc uoc cua -1 

1-y thuoc 1 -1 

ta co bang sau 

1-y   1      -1

y       0      2

x-1    -1     1

x         0      2

Answer 3

\(x+y=xy\)

\(\Rightarrow0=xy-x-y\)

\(\Rightarrow xy-x-y=0\)

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

\(\Rightarrow\left(x-1\right)\left(y-1\right)=1\)

\(\Rightarrow\left(x-1\right);\left(y-1\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)

Xét bảng

x-1	1	-1
y-1	1	-1
x	2	0
y	2	0

Vậy.........................