Question

Tìm cặp xy:

2x²-4xy+4y²-6x+9=0

Answer 1

\(2x^2-4xy+4y^2-6x+9=0\)

\(\Leftrightarrow x^2-4xy+4y^2+x^2-6x+9=0\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(x-3\right)^2=0\)

Có: \(\left(x-2y\right)^2+\left(x-3\right)^2\ge0\)

Dấu '=' xảy ra khi: \(\hept{\begin{cases}\left(x-2y\right)^2\\\left(x-3\right)^2\end{cases}}\Rightarrow\hept{\begin{cases}x-2y=0\\x-3\end{cases}}\Rightarrow\hept{\begin{cases}x-2y=0\\x=3\end{cases}}\Rightarrow\hept{\begin{cases}3-2y=0\\x=3\end{cases}}\Rightarrow\hept{\begin{cases}y=\frac{3}{2}\\x=3\end{cases}}\)

Vậy: ....