Vì \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left(y-3\right)^2\ge0\end{cases}\forall x,y\Rightarrow\left|x+1\right|+\left(y-3\right)^2\ge0\forall x,y}\)
\(\Rightarrow N=\left|x+1\right|+\left(y-3\right)^2+10\ge10\forall x,y\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x+1\right|=0\\\left(y-3\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=3\end{cases}}}\)
Vậy MinN = 10 khi x=-1,y=3
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