Ta thấy: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\forall x\\\left|y+1\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow (x+1)^2+|y-1| \geq 0 \)\(\forall x,y\)
\(\Rightarrow (x+1)^2+|y-1| -10\geq -10 \)\(\forall x,y\)
\(\Rightarrow C\ge-10\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left|y+1\right|=0\end{matrix}\right.\)\(\Leftrightarrow x=y=-1\)
Vậy với \(x=y=-1\) thì \(C_{Min}=-10\)
C = ( x + 1 )2+| y - 1 |-10
Ta xét: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left|y-1\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2+\left|y-1\right|\ge0\)
\(\Rightarrow C=\left(x+1\right)^2+\left|y-1\right|-10\ge-10\)
Vậy Min(C) = -10 khi x= -1 và y= 1
Do ( x+ 1)2>=0
/y-1/>=0
=> C>=-10
Dấu ''=" xảy ra khi \(^{\left\{{}\begin{matrix}\left(x+1\right)^2=0\\y-1=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy MinC = -10 khi x=-1 va y=1
