Question

[Hải Yến]

cho các số x y thỏa mãn đẳng thức 8x^2+14xy+8y^2 +2x-2y+2=0

Answer 1

[Nguyễn Việt Lâm]

\(8x^2+14xy+8y^2+2x-2y+2=0\)

\(\Leftrightarrow7\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)

Do \(\left\{{}\begin{matrix}7\left(x+y\right)^2\ge0\\\left(x+1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) ; \(\forall x;y\)

Nên \(7\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0;\forall x;y\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)