Ta có: \(\left(\frac{4}{7}x-1\right)^{2010}\ge0\forall x\)
\(\left(\frac{-2}{3}y+4\right)^{68}\ge0\forall y\)
\(\Rightarrow\left(\frac{4}{7}x-1\right)^{2010}+\left(\frac{-2}{3}y+4\right)^{68}\ge0\forall x,y\)
mà theo đề bài: \(\left(\frac{4}{7}x-1\right)^{2010}+\left(\frac{-2}{3}y+4\right)^{68}\le0\)
\(\Rightarrow\left(\frac{4}{7}x-1\right)^{2010}+\left(\frac{-2}{3}y+4\right)^{68}=0\)
\(\Rightarrow\left(\frac{4}{7}x-1\right)^{2010}=0;\left(\frac{-2}{3}y+4\right)^{68}=0\)
Với \(\left(\frac{4}{7}x-1\right)^{2010}=0\)
\(\Rightarrow\frac{4}{7}x-1=0\Rightarrow x=\frac{7}{4}\)
Với \(\left(\frac{-2}{3}y+4\right)^{68}=0\)
\(\Rightarrow\frac{-2}{3}y+4=0\Rightarrow y=6\)
Vậy \(\left[\begin{matrix}x=\frac{7}{4}\\y=6\end{matrix}\right.\)
Vì \(\left(\frac{4}{7}x-1\right)^{2010}\ge0,\left(\frac{-2}{3}y+4\right)^{68}\ge0\forall x\)
\(\Rightarrow\left(\frac{4}{7}x-1\right)^{2010}+\left(\frac{-2}{3}y+4\right)^{68}\ge0\forall x\)
Mà \(\left(\frac{4}{7}x-1\right)^{2010}+\left(\frac{-2}{3}y+4\right)^{68}\le0\)
\(\Rightarrow\left(\frac{4}{7}x-1\right)^{2010}=0,\left(\frac{-2}{3}y+4\right)^{68}=0\)
\(\Rightarrow\frac{4}{7}x-1=0;\frac{-2}{3}y+4=0\)
\(\Rightarrow x=\frac{7}{4},y=6\)
