- Theo đề bài ta có:
\(\left(x-178\right)^{214}+\left(1-y\right)^{356}=0\)
- Lại có:
\(\left(x-178\right)^{214}\ge0\forall x\)
\(\left(1-y\right)^{356}\ge0\forall y\)
nên \(\Rightarrow\)\(\left(x-178\right)^{214}+\left(1-y\right)^{356}\ge0\forall x;y\)
mà theo đề bài: \(\left(x-178\right)^{214}+\left(1-y\right)^{356}=0\)
nên => \(\left\{{}\begin{matrix}\left(x-178\right)^{214}=0\\\left(1-y\right)^{356}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-178=0\\1-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=178\\y=1\end{matrix}\right.\)
- Vậy x= 178
y=1
\(\left(x-178\right)^{214}+\left(1-y\right)^{356}=0\)
\(\left\{{}\begin{matrix}\left(x-178\right)^{214}\ge0\forall x\\\left(1-y\right)^{356}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-178\right)^{214}+\left(1-y\right)^{356}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-178\right)^{214}=0\\\left(1-y\right)^{356}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-178=0\Rightarrow x=178\\1-y=0\Rightarrow y=1\end{matrix}\right.\)
Vậy xảy ra khi: \(x=178;y=1\)
\(\left(x-178\right)^{214}\ge0\forall x\\ \left(1-y\right)^{356}\ge0\forall y\\ \Rightarrow\left(x-178\right)^{214}+\left(1-y\right)^{356}\ge0\forall x,y\\ \text{Mà }\left(x-178\right)^{214}+\left(1-y\right)^{356}=0\\ \Rightarrow\left\{{}\begin{matrix}\left(x-178\right)^{214}=0\\\left(1-y\right)^{356}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-178=0\\1-y=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=178\\y=1\end{matrix}\right.\)
Vậy x = 178; y = 1
\(\left(x-178\right)^{214}+\left(1-y\right)^{356}=0\Leftrightarrow\left\{{}\begin{matrix}x-178=0\\1-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=178\\y=1\end{matrix}\right.\)
