\(C=2x^2+y^2-2xy-2y+5\)
\(\Rightarrow2C=4x^2+2y^2-4xy-4y-10\)
\(2C=\left(2x\right)^2-2.2x.y+y^2+y^2-4y+4-14\)
\(2C=\left(2x-y\right)^2+\left(y-2\right)^2-14\)
Với mọi x, y ta có: \(\left(2x-y\right)^2\ge0;\left(y-2\right)^2\ge0\)
\(\Rightarrow\left(2x-y\right)^2+\left(y-2\right)^2\ge0\)
\(\Rightarrow2C=\left(2x-y\right)^2+\left(y-2\right)^2-14\ge-14\)
\(\Rightarrow C\ge\frac{-14}{2}=-7\)
Dấu bằng xảy ra khi: \(\hept{\begin{cases}2x-y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=y\\y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}2x=2\\y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)
Vậy x=1 ; y=2 thì min C = -7
HỌC TỐT <3
