Question

tìm các số nguyên x,y thỏa mãn

x² +2y²+2xy+2x-2y=0

Answer 1

Hướng dẫn:

\(x^2+2y^2+2xy+2x-2y=0\\ \Leftrightarrow\left(x^2+y^2+1+2xy+2x+2y\right)+\left(y^2-4y+4\right)=5\\ \Leftrightarrow\left(x+y+1\right)^2+\left(y-2\right)^2=1^2+2^2\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left|x+y+1\right|=\left|1\right|\\\left|y-2\right|=\left|2\right|\end{matrix}\right.\\\left\{{}\begin{matrix}\left|x+y+1\right|=\left|2\right|\\\left|y-2\right|=\left|1\right|\end{matrix}\right.\end{matrix}\right.\)