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

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

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

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Ta có:
3^12 = 9^6 = 81^3: Có số tận cùng bằng 1.
5^13: Có số tận cùng bằng 5.
7^15 = (7^16)/7 = (49^8)/7 = ((49^2)^4)/7: Có số tận cùng bằng 3.
11^2010: Có số tận cùng bằng 1.
Vậy tổng 3^12+5^13+7^15+11^2010 có số tận cùng bằng 0.

Mình biết làm mà quên mất , bây giờ mới nhớ

chữ số tận cùng là 0

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

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

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

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

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

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