Question

Tìm số nguyên x,y

\(x^2-2y^2+xy+x-y-5=0\)

Answer 1

\(x^2-2y^2+xy+x-y-5=0\)

\(\Leftrightarrow x^2-y^2+xy-y^2+\left(x-y\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)+y\left(x-y\right)+\left(x-y\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x+y+y+1\right)=5\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y+1\right)=5\)

\(\Leftrightarrow\left(x-y\right);\left(x+2y+1\right)\in U\left(5\right)=\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(2;1\right);\left(-4;1\right)\right\}\)