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\(C=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{1^{1010}.2^{2010}+2^{2010}.2^{2010}+2^{2010}.3^{2010}+...+2^{2010}.10^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{\left(1^{2010}+2^{2010}+3^{2010}+...+10^{2010}\right)+2^{2010}.2^{2010}.2^{2010}...2^{2010}}\)
\(=\dfrac{1}{2^{2010}+2^{2010}+2^{2010}+...+2^{2010}}\)
\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+....+20^{2010}}\\ =\dfrac{1^{2010}+2^{2010}+...+10^{2010}}{2^{2010}\left(1^{2010}+2^{2010}+...+10^{2010}\right)}\\ =\dfrac{1}{2^{2010}}\)
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=-3^{2010}\)
Theo anh thì:
M=(1+2010)+(2010^2+2010^3)+(2010^4+2010^5)+(2010^6+2010^7)
M=(1+2010)+2010^2(1+2010)+2010^4(1+2010)+2010^6(1+2010)
M=2011(2010^2+1010^4+2010^6) Vậy M chia hết cho 2011 vì trong 1 tích chỉ cần có 1 thừa số chia hết cho 1 số thì cả tích đó chia hết cho số đó.
Rút gọn:
32020 - 62010 + 92010 - 122010 + 152010 - 182010
-1 + 22010 - 32010 + 42010 - 52010 + 62010
Đặt \(x=2009\)
\(A=2009^8-2010\cdot2009^7+2010\cdot2009^6-2010\cdot2009^5+...+2010\cdot2009^0\)
\(\Leftrightarrow A=x^8-\left(x+1\right)x^7+\left(x+1\right)x^6-\left(x+1\right)x^5+...+\left(x+1\right)x^0\\ \Leftrightarrow A=x^8-x^8-x^7+x^7+x^6-x^6-x^5+...-x^2-x^1+x^1+x^0\)
\(\Leftrightarrow A=x^0\\ \Leftrightarrow A=1\)