Question

tim gtnn cua P=(|x-3|+2)^2 + |y+3| + 2007

Answer 1

\(P=\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2017\) 

Có: \(\left(\left|x-3\right|+2\right)^2\ge4\)

\(\left|y+3\right|\ge0\)

\(\Rightarrow\left(\left|x-3\right|+2\right)^2+\left|y+3\right|+2017\ge2021\)

\(\Leftrightarrow\hept{\begin{cases}\left(\left|x-3\right|+2\right)^2=4\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x-3=0\\y=-3\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

Vậy: \(Min_P=2021\) tại \(\hept{\begin{cases}x=3\\y=-3\end{cases}}\)