Ko là mà muốn  có ăn thì chỉ có ăn cứt thôi!

\(C=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)

\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{1^{1010}.2^{2010}+2^{2010}.2^{2010}+2^{2010}.3^{2010}+...+2^{2010}.10^{2010}}\)

\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{\left(1^{2010}+2^{2010}+3^{2010}+...+10^{2010}\right)+2^{2010}.2^{2010}.2^{2010}...2^{2010}}\)

\(=\dfrac{1}{2^{2010}+2^{2010}+2^{2010}+...+2^{2010}}\)

\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+....+20^{2010}}\\ =\dfrac{1^{2010}+2^{2010}+...+10^{2010}}{2^{2010}\left(1^{2010}+2^{2010}+...+10^{2010}\right)}\\ =\dfrac{1}{2^{2010}}\)

Theo bài ra, ta có:

\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+....+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+....+20^{2010}}\)

\(\Rightarrow G=\dfrac{1^{2010}+2^{2010}+3^{2010}+....+10^{2010}}{2^{2010}\left(1^{1010}+2^{2010}+3^{2010}+....+10^{2010}\right)}\)

\(\Rightarrow G=\dfrac{1}{2^{2010}}\)

Vậy \(G=\dfrac{1}{2^{2010}}\)

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\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)

\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)

\(P=-3^{2010}\)

A=(1+2010)+2010 mũ 2+2010 mũ 3 +...+2010 mũ 6 + 2010 mũ 7

A=2011+2010 mũ 2(1+2010)+...+2010 mũ 6(1+2010)

A=2011+2010 mũ 2.2011+...2010 mũ 6.2011

A=2011(1+2010+...+2010 mũ 6)chia hết cho 2011