Ta có: \(A=x^2-2xy+2y^2-4y+5\)
\(\Leftrightarrow A=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)
\(\Leftrightarrow A=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu "=" xảy ra khi: \(x=y=2\)
Vậy ...
Ta có:
\(A=x^2-2xy+2y^2-4y+5\)
\(A=\left(x^2-2xy+y^2\right)+\left(y^2-4y+4\right)+1\)
\(A=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu " = " xảy ra khi \(x=y=2\)
Rất vui vì giúp đc bạn !!!
\(A=x^2-2xy+2y^2-4y+5\)
\(=x^2-2xy+y^2+y^2-4y+4+1\)
\(=\left(x-y\right)^2+\left(y-2\right)^2+1\ge1\)
Dấu \("="\)xảy ra\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2=0\\\left(x-2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x-y=0\\x-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=2\end{cases}\Rightarrow}x=y=2}\)
Vậy \(GTNN\)của\(A\)là \(1\Leftrightarrow x=y=2\)
