Question

[Nguyễn Tuấn Anh]

Tìm x, y nguyên thỏa mãn: x² + 2xy +  7( x + y) + 2y² + 10 = 0

Answer 1

[bui van trong]

\(x^2+2xy+7\left(x+y\right)+2y^2+10=0\)0

\(< =>\left(x^2+2xy+y^2\right)+7\left(x+y\right)+y^2+10=0\)

\(< =>\left(x+y\right)^2+7\left(x+y\right)+y^2+10=0\)

Đặt a=x+y ta có

\(a^2+7a+10+y^2=0\)

\(< =>a^2+7a+\frac{49}{4}-\frac{9}{4}+y^2=0\)

\(< =>\left(a+\frac{7}{2}\right)^2+y^2=\frac{9}{4}\)

Vì \(\frac{9}{4}\)=\(0+\frac{9}{4}\)và \(a+\frac{7}{2}>=y\)nên \(\hept{\begin{cases}x+y+\frac{7}{2}=\frac{3}{2}\\y=0\end{cases}}\)\(=>\hept{\begin{cases}y=0\\x=-2\end{cases}}\)