Question

Tìm các số thực x,y thoả mãn
2x² + y² - 2xy - 8x + 16 = 0

Answer 1

\(2x^2+y^2-2xy-8x+16=0\)

\(\Leftrightarrow\left(x^2-8x+16\right)+\left(x^2-2xy+y^2\right)=0\)

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

Do: \(\left\{{}\begin{matrix}\left(x-4\right)^2\ge0\\\left(x-y\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-4\right)^2+\left(x-y\right)^2\ge0\)

Mặt khác: \(\left(x-4\right)^2+\left(x-y\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\x-y=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=4\)

Vậy: ...