2010<y*7+1996<2020
H24
Những câu hỏi liên quan
1^2010+2^2010+3^2010+...+10^2010 / 2^2010+4^2010+6^2010+...+20^2010
Rút gọn
A=\(\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
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rút gọn biểu thức: P=\(\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
Rút gọn biểu thức:
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{3^{2010}-6^{2010}+9^{2010}-12^{2010}+15^{2010}-18^{2010}}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=\frac{-3^{2010}.\left(-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}\right)}{-1+2^{2010}-3^{2010}+4^{2010}-5^{2010}+6^{2010}}\)
\(P=-3^{2010}\)
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Giúp mình với : So sánh : A = 2010^2010 -3 /2010^2010+4 B = 2010^2010-4 / 2010^2010
Tính giúp mình câu tính nhanh này nha.2010+2010+2010+2010+2010-2010×4
Xem chi tiết
2010x5-2010x4
=2010x(5-4)
=2010x1
=2010
Chúc bạn học tốt nha
Rú gọn biểu thức G = \(\frac{1^{2010}+2^{2010}+3^{2010}+....................+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+.....................+20^{2010}}\)
chung minh 1+2010^1+2010^2+2010^3+2010^4+2010^5+2010^6+2010^7 chia het cho11 help me
rút gọn biểu thức :
G = \(\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)
\(C=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+6^{2010}+...+20^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{1^{1010}.2^{2010}+2^{2010}.2^{2010}+2^{2010}.3^{2010}+...+2^{2010}.10^{2010}}\)
\(=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{\left(1^{2010}+2^{2010}+3^{2010}+...+10^{2010}\right)+2^{2010}.2^{2010}.2^{2010}...2^{2010}}\)
\(=\dfrac{1}{2^{2010}+2^{2010}+2^{2010}+...+2^{2010}}\)
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\(G=\dfrac{1^{2010}+2^{2010}+3^{2010}+...+10^{2010}}{2^{2010}+4^{2010}+....+20^{2010}}\\ =\dfrac{1^{2010}+2^{2010}+...+10^{2010}}{2^{2010}\left(1^{2010}+2^{2010}+...+10^{2010}\right)}\\ =\dfrac{1}{2^{2010}}\)
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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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Xem thêm câu trả lời
Choa,b,c,d khác 0:
\(\frac{x^{2010}+y^{2010}z^{2010}+t^{2010}}{a^2+b^2+c^2+d^2}=\frac{x^{2010}}{a^2}+\frac{y^{2010}}{b^2}+\frac{z^{2010}}{c^2}+\frac{t^{2010}}{d^2}\)
Tính\(T=x^{2010}+y^{2010}+z^{2010}+t^{2010}\)