Question

Answer 1

\(x^2+xy+x+y=1\)

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

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

Nhưng x,y ϵ N nên:

\(x+1=1\)

\(\Rightarrow x=1-1\)

\(\Rightarrow x=0\)

\(x+y=1\)

\(\Rightarrow0+y=1\)

\(\Rightarrow y=1\)

Vậy: ...

Answer 2

\(x^2+xy+x+y=1\) \(\left(x;y\in N\right)\)

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

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

\(\Rightarrow\left(x+1\right);\left(x+y\right)\in U\left(1\right)=\left\{1\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;1\right)\right\}\)