a) \(A=\left(x-1\right)^2+\left(y+3\right)^2+2010\)
Vì \(\left(x-1\right)^2\ge0\forall x;\left(y+3\right)^2\ge0\forall y\)
\(\Rightarrow A\ge2010\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}}\)
Vậy Amin = 2010 <=> x = 1; y = -3
b) tương tự
a, Nhận xét : \(\left(x-1\right)^2\ge0\text{ với}\forall x\)
\(\left(y+3\right)^2\ge0\text{ với}\forall y\)
=> \(\left(x-1\right)^2+\left(y+3\right)^2\ge0\text{ với}\forall x,y\)
=> \(\left(x-1\right)^2+\left(y+3\right)^2+2010\ge2010\)
=> Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
=> \(A_{min}=2010\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
b, \(B=\left(x-3\right)^2+\left(y-1\right)^2+5\)
Nhận xét : \(\left(x-3\right)^2\ge0\)với \(\forall x\)
\(\left(y-1\right)^2\ge0\)với \(\forall y\)
=> \(\left(x-3\right)^2+\left(y-1\right)^2\ge0\)
=> \(\left(x-3\right)^2+\left(y-1\right)^2+5\ge5\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3=0\\y-1=0\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=1\end{cases}}\)
Vậy \(B_{min}=5\Leftrightarrow\hept{\begin{cases}x=3\\y=1\end{cases}}\)
