\(\left(x+2023\right)^{40}+\left(y+2022\right)^{10}=0\)
Ta thấy: \(\left(x+2023\right)^{40}\ge0\forall x\)
\(\left(y+2022\right)^{10}\ge0\forall x\)
\(\Rightarrow\left(x+2023\right)^{40}+\left(y+2022\right)^{10}\ge0\forall x\)
Mặt khác: \(\left(x+2023\right)^{40}+\left(y+2022\right)^{10}=0\)
nên: \(\left\{{}\begin{matrix}\left(x+2023\right)^{40}=0\\\left(y+2022\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2023=0\\y+2022=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2023\\y=-2022\end{matrix}\right.\)
#Toru
Ta có:
\(\left(x+2023\right)^{40}+\left(y+2022\right)^{10}=0\)
Mà: \(\left\{{}\begin{matrix}\left(x+2023\right)^{40}\ge0\forall x\\\left(y+2022\right)^{10}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x+2023\right)^{40}+\left(y+2022\right)^{10}\ge0\forall x,y\)
Dấu "=" xảy ra cũng là nghiệm của phương trình
\(\left\{{}\begin{matrix}\left(x+2023\right)^{40}=0\\\left(y+2022\right)^{10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2023=0\\x+2022=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2023\\y=-2022\end{matrix}\right.\)
Vậy: ....
\(\left(x+2023\right)^{40}+\left(y+2022\right)^{10}=0\left(1\right)\)
Vì \(\left\{{}\begin{matrix}\left(x+2023\right)^{40}\ge0,\forall x\in R\\\left(y+2022\right)^{10}\ge0,\forall y\in R\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x+2023\right)^{40}=0\\\left(y+2022\right)^{10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2023=0\\y+2022^{ }=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2023\\y=-2022\end{matrix}\right.\)
