Question

Tìm các số tự nhiên x, y thỏa mãn : x² + 3y² + 4xy + 4x + 10y - 12 = 0

Answer 1

\(x^2+3y^2+4xy+4x+10y-12=0\)

\(\Leftrightarrow\left(x^2+4y^2+4+4xy+4x+8y\right)-\left(y^2-2y+1\right)-17=0\)

\(\Leftrightarrow\left(x+2y+2\right)^2-\left(y-1\right)^2=17\)

\(\Leftrightarrow\left(x+3y+1\right)\left(x+y+3\right)=17\)

. . . . . . . . . . . . .. . . . . . . . . . . . .