Question

[Haruno :3]

Tìm x,y thoả mãn: y²+2xy-12x+4(x+y)+2x²+40=0

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+4\left(x+y\right)+4+\left(x^2-12x+36\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+4\left(x+y\right)+4+\left(x-6\right)^2=0\)

\(\Leftrightarrow\left(x+y+2\right)^2+\left(x-6\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=-8\end{matrix}\right.\)

Answer 2

[Nguyen Minh Hieu]

\(y^2+2xy-12x+4\left(x+y\right)+2x^2+40=0\\ \Leftrightarrow\left[\left(x^2+2xy+y^2\right)+4\left(x+y\right)+4\right]+\left(x^2-12x+36\right)=0\\ \Leftrightarrow\left(x+y+2\right)^2+\left(x-6\right)^2=0\)

Vì \(\left\{{}\begin{matrix}\left(x+y+2\right)^2\ge0\forall x,y\\\left(x-6\right)^2\ge0\forall x\end{matrix}\right.\) 

Nên \(\left(x+y+2\right)^2+\left(x-6\right)^2\ge0\forall x,y\)

Dấu"=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x+y+2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-8\\x=6\end{matrix}\right.\)

Vậy x = 6 và y = -8