\(\text{Do}\left(\frac{1}{3}-2x\right)^{120}\ge0\text{ với }\forall x\in Q\)
\(\left(3y+x\right)^{104}\ge0\text{ với }\forall x,y\in Q\)
\(\Rightarrow\text{}\left(\frac{1}{3}-2x\right)^{120}+\left(3y+x^{104}\right)\ge0\)
\(\text{Mà }\left(\frac{1}{3}-2x\right)^{120}+\left(3y+x^{104}\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\frac{1}{3}-2x\right)^{120}=0\\\left(3y-x\right)^{104}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\frac{1}{3}\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\3y-\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\3y=\frac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{6}\\y=\frac{1}{18}\end{matrix}\right.\)
Vậy \(x=\frac{1}{6},y=\frac{1}{18}\)
