Vì \(\left(2x-1\right)^{2008}\ge0;\left(y+3.1\right)^{2008}\ge0\)
\(\Rightarrow\left(2x-1\right)^{2008}+\left(y+3.1\right)^{2008}\ge0\)
Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(2x-1\right)^{2008}=0\\\left(y+3.1\right)^{2008}=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-1=0\\y+3.1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}}\)
\(\left(2x-1^{ }\right)^{2008}\)+ \(\left(y+3.1\right)^{2008}\)\(=0\)
ta có : \(\left(2x-1\right)^{2008}>=0\)
\(\left(y+3.1^{ }\right)^{2008}>=0\)
TH1 :\(\left(2x-1\right)=0\)=>\(2x=1\)=> \(x=0.5\)
TH2 : \(y+3.1=0\)=> \(y+3=0\)=>\(y=-3\)
Ta có \(x+y=0.5+-3=-2.5\)
cho mk nha
