Do \(\left\{{}\begin{matrix}a^{2008}\ge0\\b^{2008}\ge0\\c^{2008}\ge0\\a^{2008}+b^{2008}+c^{2008}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^{2008}\le1\\b^{2008}\le1\\c^{2008}\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\)

\(\Rightarrow a^{2009}+b^{2009}+c^{2009}\le a^{2008}+b^{2008}+c^{2008}\)

\(\Rightarrow a^{2009}+b^{2009}+c^{2009}\le1\)

Dấu "=" xảy ra khi và chỉ khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị

Khi đó \(a^{2007}+b^{2008}+c^{2009}+2020=1+2020=2021\)

\(a-b+c+d=\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)

\(=\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)\)

\(=1-\frac{2}{2008}\)

\(=\frac{1003}{1004}\)

1003/1004

\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}=\frac{1003}{1004}\)

ai k mình mình k lại,ok

\(\frac{1003}{1004}\)

Ta có:

\(a^{2006}+a^{2008}+b^{2006}+b^{2008}\ge2\left(a^{2007}+b^{2007}\right)\)

Dấu = xảy ra khi \(a=b=1\)

\(\Rightarrow S=a^{2009}+b^{2009}=2\)

a-b+c+d=\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}=\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)=1-\frac{2}{2008}=\frac{2006}{2008}=\frac{1003}{1004}\)

\(a-b+c+d=\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)

\(=\left(\frac{2008}{2009}+\frac{1}{2009}\right)+\left(\frac{2007}{2008}-\frac{2009}{2008}\right)=\frac{2009}{2009}+\frac{-2}{2008}\)

\(=1+\frac{-1}{1004}=\frac{1004}{1004}+\frac{-1}{1004}=\frac{1003}{1004}\)

Ta có: \((a^{2007}+b^{2007})\left(a+b\right)-\left(a^{2006}+b^{2006}\right)ab\)

\(=\left(a^{2008}+a^{2007}b+ab^{2007}+b^{2008}\right)-\left(a^{2007}b+ab^{2007}\right)\)

\(=a^{2008}+b^{2008}\)

Mà: \(a^{2006}+b^{2006}=a^{2007}+b^{2007}=a^{2008}+b^{2008}\)    ( * )

\(\Rightarrow\left(a^{2008}+b^{2008}\right)\left(a+b\right)-\left(a^{2008}+b^{2008}\right)ab=a^{2008}+b^{2008}\)

\(\Leftrightarrow\left(a^{2008}+b^{2008}\right)\left(a+b-ab\right)=a^{2008}+b^{2008}\)

\(\Leftrightarrow a+b-ab=1\)

\(\Leftrightarrow\left(a-1\right)-b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(1-b\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a=1\\b=1\end{cases}}\)

thay vào (*) ta tính dc: 

a=1 thì\(\orbr{\begin{cases}b=1\\b=0\end{cases}}\)                   b=1 thì \(\orbr{\begin{cases}a=1\\a=0\end{cases}}\)

mặt khác a, b dương => a=1, b=1

Khi đó:   \(a^{2009}+b^{2009}=1+1=2\)

Ta có : \(a^{2006}+b^{2016}=a^{2007}+b^{2007}=a^{2008}+b^{2008}\)

\(\Leftrightarrow\orbr{\begin{cases}a^{2006}+b^{2006}-\left(a^{2007}+a^{2007}\right)=0\left(1\right)\\a^{2008}+b^{2008}-\left(a^{2007}+b^{2007}\right)=0\left(2\right)\end{cases}}\) 

Cộng (1) với (2)  => \(a^{2008}+b^{2008}-2\left(a^{2007}+b^{2007}\right)+a^{2006}+b^{2006}=0\)

\(\Leftrightarrow a^{2008}-2a^{2007}+a^{2006}+b^{2008}-2b^{2007}+b^{2006}\)

\(\Leftrightarrow a^{2006}\left(a^2-2a+1\right)+b^{2006}\left(b^2-2b+1\right)=0\)

\(\Leftrightarrow a^{2006}\left(a-1\right)^2+b^{2006}\left(b-1\right)^2=0\) (*) 

Vì a , b > 0 và : \(\left(a-1\right)^2\ge0\forall a\) ; \(\left(b-1\right)^2\ge0\forall b\)

Nên : phương trình (*) <=> \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\end{cases}\Leftrightarrow a=b=1}}\)

Vậy \(S=a^{2009}+b^{2009}=1+1=2\)