Question

Cho các số x,y thỏa mãn đẳng thức: x² + xy + y² + x - y + 1 = 0

Tính \(\left(x+y\right)^{30}+\left(x+2\right)^{12}+\left(y-1\right)^{2017}\).

Answer 1

\(x^2+xy+y^2+x-y+1=0\)

\(\Leftrightarrow2x^2+2xy+2y^2+2x-2y+2=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\) (*)

Vì \(\left(x+y\right)^2\ge0;\left(x+1\right)^2\ge0;\left(y-1\right)^2\ge0\)

(*) \(\Leftrightarrow\left(x+y\right)^2=0;\left(x+1\right)^2=0;\left(y-1\right)^2=0\)

\(\Leftrightarrow x+y=0;x+1=0;y-1=0\)

\(\Rightarrow x+2=1\)

\(\Rightarrow\left(x+y\right)^{30}+\left(x+2\right)^{12}+\left(y-1\right)^{2017}=0+1+0=1\)