sữa đề ta có : \(\left(\dfrac{4}{7}x-1\right)^{2010}+\left(-\dfrac{2}{3}y+4\right)^{68}\le0\)
mà : \(\left\{{}\begin{matrix}\left(\dfrac{4}{7}x-1\right)^{2010}\ge0\\\left(-\dfrac{2}{3}y+4\right)^{68}\ge0\end{matrix}\right.\) \(\Rightarrow\) : \(\left(\dfrac{4}{7}x-1\right)^{2010}+\left(-\dfrac{2}{3}y+4\right)^{68}\ge0\)
mà \(\left(\dfrac{4}{7}x-1\right)^{2010}+\left(-\dfrac{2}{3}y+4\right)^{68}\le0\)
vậy \(\Rightarrow\) \(\left(\dfrac{4}{7}x-1\right)^{2010}+\left(-\dfrac{2}{3}y+4\right)^{68}=0\)
\(\left\{{}\begin{matrix}\left(\dfrac{4}{7}x-1\right)^{2010}=0\\\left(-\dfrac{2}{3}y+4\right)^{68}=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{4}{7}x-1=0\\-\dfrac{2}{3}y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{4}{7}x=1\\-\dfrac{2}{3}y=-4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{7}{4}\\y=6\end{matrix}\right.\)
vậy \(x=\dfrac{7}{4};y=6\)
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