x+y+xy=11

=> x(y+1)+y=11

=> x(y+1)+y+1=12

=> (y+1)(x+1)=12=1.12;2.6;3.4;4.3;6.2;12.1

Vậy: (x;y)= (11;0);(0;11);(5;1);(1;5);(2;3);(3;2)


 

Còn x+6=y(x-1)

Hưỡng dẫn: Đổi vế x từ x+6 sau đó trừ cả vế với 1 nhóm lại ta được hai thừa số

Sau đó lập thành tích.

Cậu biết giải rồi mà còn đăng bài làm gì nữa? Để khoe hay sao vậy?

a)Do x,y là STN mà xy=6=1.6=2.3

=>(x;y)={(1;6);(6;1);(2;3);(3;2)}

b)Do x,y là STN mà xy=40=1.40=2.20=4.10=8.5

=>(x;y)={(1;40);(40;1);(2;20);(20;2);(4;10);(10;4);(8;5);(5;8)}

x(y+2)+3y =6

=>x(y+3)+3y+9=15

=>x(y+3)+3(y+3)=15

=>(x+3)(y+3)=15

mả .....=......=>ta co bang sau

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xy+x+y=6

<=>x.(y+1)+y+1=7

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

 Rồi liệt kê ra

Không có giá trị x,y phù hợp

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