\(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)
Ta có: \(\left(3x-\frac{1}{5}\right)^{2020}\ge0\forall x\)
\(\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}\ge0\forall y\)
Mà \(\left(3x-\frac{1}{5}\right)^{2020}+\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(3x-\frac{1}{5}\right)^{2020}=0\\\left(\frac{2}{5}\cdot y+\frac{4}{7}\right)^{2020}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x-\frac{1}{5}=0\\\frac{2}{5}\cdot y+\frac{4}{7}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=\frac{1}{5}\\\frac{2}{5}\cdot y=\frac{-4}{7}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{15}\\y=\frac{-10}{7}\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(\frac{1}{15};\frac{-10}{7}\right)\)
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