\(\left(x-2\right)^{20}\ge0\forall x\\ \left(y+1\right)^{30}\ge0\forall x\\ \Rightarrow\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\forall x\)
Mà \(\left(x-2\right)^{20}+\left(y+1\right)^{30}=0\)
Để thỏa mãn điều kiện thì \(\left\{{}\begin{matrix}\left(x+2\right)^{20}=0\\\left(y+1\right)^{30}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+2=0\\y+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
Vậy ...
ta có : \(\left(x-2\right)^{20}\ge0\) với mọi x
và \(\left(y+1\right)^{30}\ge0\) với mọi y
\(\Rightarrow\) \(\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\) với mọi giá trị của x ; y
mà \(\left(x-2\right)^{20}+\left(y+1\right)^{30}\le0\)
\(\Rightarrow\) \(\left(x-2\right)^{20}+\left(y+1\right)^{30}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{20}=0\\\left(y+1\right)^{30}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\) vậy \(x=2;y=-1\)
\(\left(x-2\right)^{20}+\left(y+1\right)^{30}\le0\)
\(\left\{{}\begin{matrix}\left(x-2\right)^{20}\ge0\forall x\\\left(y+1\right)^{30}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^{20}+\left(y+1\right)^{30}\ge0\\\left(x-2\right)^{20}+\left(y+1\right)^{30}\le0\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^{20}+\left(y+1\right)^{30}=0\)
Daqaus "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-2\right)^{20}=0\Rightarrow x=2\\\left(y+1\right)^{30}=0\Rightarrow y=-1\end{matrix}\right.\)
Vậy xảy ra khi \(x=2;y=-1\)
