Vì \(\left(2x+\frac{1}{3}\right)^{100}\ge0\left(\forall x\right)\); \(\left(3y-\frac{1}{4}\right)^{100}\ge0\left(\forall y\right)\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{100}+\left(3y-\frac{1}{4}\right)^{100}\ge0\left(\forall x,y\right)\)
mà \(\left(2x+\frac{1}{3}\right)^{100}+\left(3y-\frac{1}{4}\right)^{100}\le0\)( giả thiết )
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{100}+\left(3y-\frac{1}{4}\right)^{100}=0\)\(\Leftrightarrow\hept{\begin{cases}2x+\frac{1}{3}=0\\3y-\frac{1}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=\frac{-1}{3}\\3y=\frac{1}{4}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{6}\\y=\frac{1}{12}\end{cases}}\)
Vậy \(x=\frac{-1}{6}\)và \(y=\frac{1}{12}\)
