Ta có: \(\hept{\begin{cases}\left(x+\frac{1}{2}\right)^{100}\ge0;\forall x,y\\|7-\frac{1}{3}y|\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(x+\frac{1}{2}\right)^{100}+|7-3y|\ge0;\forall x,y\)
Do đó \(\left(x+\frac{1}{2}\right)^{100}+|7-3y|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{2}\right)^{100}=0\\|7-\frac{1}{3}y|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\7-\frac{1}{3}y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{7}{3}\end{cases}}\)
Vậy ...
Nhầm nhé \(y=21\)
