\(\left(x+1\right)^{2006}+\left(y-1\right)^{2008}=0\)
\(\left\{{}\begin{matrix}\left(x+1\right)^{2006}\ge0\\\left(y-1\right)^{2008}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^{2006}+\left(y-1\right)^{2008}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x+1\right)^{2006}=0\\\left(y-1\right)^{2008}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Thay vào C ta có:
\(C=5.\left(-1\right)^{10}-1^{15}+2007\)
\(=5-1+2007=2011\)
(x+1)2006+(y-1)2008=0
=> (x+1)2006=(y-1)2008=0
=>x+1=y-1=0
=>x=-1 và y=1
C=5x10-y15+2007=5.(-1)10-115+2007=2011
