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20 tháng 7 2017
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
2 tháng 2 2016
đặt S=1+2+2^2+2^3+...+2^1007
2S=2+2^2+2^3+2^4+...+2^1008
2S-S=2+2^2+2^3+2^4+...+2^1008-1-2-2^2-2^3-..-2^1007
S=2^1008-1
chọn a
19 tháng 8 2015
C = 1.2.3.4....2011
D = 1007/2 . 1008/2 . 1009/2.....2012/2
D = (1007.1008.1009.....2012) : (2.2.2.2........2) (có 2012 - 1007 + 1 = 1006 số 2 )
D = (1007.1008.1009....503 .2.2) : 21004
MÀ (4.503.1007.1008.....2011) < (1.2.3.....2011)
Vậy c > D
4 tháng 4 2015
Ta có:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\)
=>\(\left(\frac{A}{B}\right)^{2013}\)=(\(\frac{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}^{ }\))2013=12013=1
LD
13 tháng 2 2022
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2015.2016}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1008}\right)\)
\(A=\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2015}+\frac{1}{2016}\)
\(\Rightarrow B-A=\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}\right)-\left(\frac{1}{1009}+\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2016}\right)\)
\(\Rightarrow B-A=\frac{1}{1008}\)
DD
2
29 tháng 1 2016
đặt tổng trên là A
=>2A=2+2^2+2^3+...+2^1008
=>2A-A=2^1008-1
=>A=2^1008-1
ko có đáp án nào ở trên