Question

[Vũ Thành Đạt]

tìm các cặp số nguyên x y sao cho:

(x+1)^2 +(y+1)^2+(x-y)^2=2

Answer 1

[❥︵₣σrεvëɾ™]

\(^{\text{(x+1+y+1+x+y)}^2}\)=2

\(^{\text{(x+1+y+1+x+y)}^2}\) =\(^{2^2}\)

x+1+y+1+x+y=2

(x+x)+(y+y)+(1+1)=2

x.2+y.2+2=2

2.(x+y+1)=2

x+y+1=2:2

x+y+1=1

x+y=1-1

x+y=0

=>x;y=0