Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left|y+2\right|\ge0\forall y\end{cases}}\)
=> \(\left(x-1\right)^2+\left|y+2\right|\ge0\forall x,y\)
Dấu " = " xảy ra khi và chỉ khi \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left|y+2\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy GTNN là 0 khi x = 1,y = -2
<=> x = 1,y = -2
Bài giải
\(\left(x-1\right)^2+\left|y+2\right|=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left|y+2\right|\ge\forall x\end{cases}}\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left|y+2\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }\left(x\text{ ; }y\right)=\left(1\text{ ; }-2\right)\)
Ta có: \(\left(x-1\right)^2+\left|y+2\right|=0\)
\(\Leftrightarrow\left(x-1\right)^2=-\left|y+2\right|\)
Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\-\left|y+2\right|\le0\end{cases}\left(\forall x,y\right)}\) nên dấu "=" xảy ra khi:
\(\left(x-1\right)^2=-\left|y+2\right|=0\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Ta có:\(\orbr{\begin{cases}\left(x-1\right)^2\\\left|y+2\right|\end{cases}\ge0\forall x,y}\)
\(\Rightarrow\left(x-1\right)^2+\left|y+2\right|\ge0\forall x,y\)
Dấu"="xảy ra khi \(\orbr{\begin{cases}\left(x-1\right)^2=0\\\left|y+2\right|=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\y=-2\end{cases}}}\)
