TÍnh
1*2016+2*2009+3*2008+...+2010*1
(1+2+3+...+2010)+(1+2+3+...+2009)+...+(1+2)+1
Những câu hỏi liên quan
a) 2010/1+2009/2+2008/3+ ... +1/2010+2010 : 1+1/2+1/3+ ... +1/2010=
b) 1/2011+1/2010+1/2009+ ... +1/3+1/2 : 2010/1+2009/2+2008/3+ ... +1/2010=
\(\dfrac{x+1}{2010}+\dfrac{x+2}{2009}+\dfrac{x-3}{2008}+...+\dfrac{x-2009}{2}+\dfrac{x-2010}{1}=-2010\)
\(\Leftrightarrow\dfrac{x+1}{2010}+1+\dfrac{x+2}{2009}+1+...+\dfrac{x+2009}{2}+1+\dfrac{x+2010}{1}+1=0\)
=>x+2011=0
hay x=-2011
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ai đó giúp mk với mk xin chân thành cảm ơn! a=(2010+2010/2+2009/3+2008/4+...+1/2011/ 1/2+1/3+...+1/2011) / (1/2+1/3+1/4+1/5+...+1/2009+1/2010+1/2011)
tính tổng sau :\(c=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\)\(\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)
\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)
\(=\frac{1}{5}+\frac{2}{3}\)
\(=\frac{13}{15}\)
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Tính nhanh
1* 2010+2*2009+3*2008+....+2010*1 phần [1+2+3+...+2010]+[1+2+3+...+2009]+...+[1+2] +1
dấu * này là nhân
lầm hộ mình nhé
Cho A = 1/2001+2/2009+3/2008+........2009/+ 2010/1, B = 1+1/2+1/3+1/4+1/5+1/6+.......1/2010+1/2011. Tính A/B
Tính: A = (1- 1/2)(1-1/3)(1-1/4)...(1-1/2016)(1-1/2017)
S= 2^2010 - 2^2009 - 2^2008 - ... - 2 - 1
Câu 1:
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(A=\frac{1}{2017}\)
Vậy ..............................
Phần giống nhau là gạch ý!
Câu 2
\(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Rightarrow S=2^{2010}-\left(2^{2009}+2^{2008}+...+2+1\right)\)
Đặt \(Q=2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow2Q=2^{2010}+2^{2009}+...+4+2\)
\(\Rightarrow2Q-Q=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)
\(\Rightarrow Q=2^{2010}-1\)
\(\Rightarrow S=2^{2010}-\left(2^{2010}-1\right)\)
\(\Rightarrow S=2^{2010}-2^{2010}+1\)
\(\Rightarrow S=1\)
Vậy .........................
b) \(S=2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Rightarrow2S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)
\(\Rightarrow2S-S=\left(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\right)-\left(2^{2010}-2^{2009}-2^{2008}-...-2-1\right)\)
\(\Rightarrow S=2^{2011}-2^{2010}-2^{2009}-...-2^2-2-2^{2010}+2^{2009}+2^{2008}+...+2+1\)
\(\Rightarrow S=2^{2011}-2^{2010}-2^{2010}+1\)
\(\Rightarrow S=2^{2011}-2.2^{2010}+1\)
\(\Rightarrow S=2^{2011}-2^{2011}+1\)
\(\Rightarrow S=0+1\)
\(\Rightarrow S=1.\)
Vậy \(S=1.\)
Chúc bạn học tốt!
Tính nhanh:
a,1/2009+2/2009+3/2009+......2008/2009
b,2010*2010*20092009-2009*2009*20102010/2009*20052005
a, \(\frac{1}{2009}+\frac{2}{2009}+...+\frac{2008}{2009}\\ \frac{\left(1+2008\right)\cdot2008\div2}{2009}=\frac{2017036}{2009}\)
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Tính tổng:
a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)
b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)
a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)
S1=-1.(2010:2)
S1=-1005
b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)
S2=-1.(2010:2)
S2=-1.1005
S2=-1005
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