Có :
\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\)
Mà theo đề bài : \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)