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PT
2
30 tháng 12 2017
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2015.2016}=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{1}{2}-\dfrac{1}{2016.2}< \dfrac{1}{2}\left(đpcm\right)\)
NT
30 tháng 12 2017
\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2015.2017}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2015.2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2017}\right)\\ < \dfrac{1}{2}.1=\dfrac{1}{2}\)
DT
1
18 tháng 9 2016
S = 4/2.3 . 10/3.4 ..........9898/ 99.100
S= 1.4/2.3 . 2.5/3.4 . 3.6/4.5 .......98.101/99.100
S = (1.2.3...98).(4.5.6...100).1014/(2.3.4...98).99.(4.5.6...100).3
S=101/99.3
S=101.297
PM
0
HH
3
16 tháng 7 2019
\(D=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)
\(=2015-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=2015-\left(1-\frac{1}{2016}\right)\)
\(=2015-\frac{2015}{2016}\)
TO LẮM
HT
17 tháng 6 2015
\(\frac{x}{2}+\frac{x}{2.3}+\frac{x}{3.4}+.....+\frac{x}{2015.2016}=\frac{2015}{4032}\)
\(x.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\left(1-\frac{1}{2016}\right)=\frac{2015}{4032}\)
\(x.\frac{2015}{2016}=\frac{2014}{4032}\)
\(x=\frac{2015}{4032}:\frac{2015}{2016}\)
\(x=\frac{1}{2}\)
17 tháng 6 2015
\(=\frac{x}{1}-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2015}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{x}{1}-\frac{x}{2016}=\frac{2015}{4023}\)
\(=\frac{2015}{2016}x=\frac{2015}{4023}\)
=> x = \(\frac{2015}{4023}\cdot\frac{2016}{2015}\)= 2016/4023
NC
1
HP
21 tháng 2 2016
2x+4x+6x+...+2014x=2015.2016
=>(2+4+6+...+2014).x=2015.2016
tổng trong ngoặc có:(2014-2):2+1=1007(số hạng)
=>tổng= (2014+2).1007:2=1015056
=>1015056.x=4062240
=>x=4030/1007
Ta có : \(x=\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{1}{1008.2015}\)
\(y=\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Vì \(\frac{1}{1008.2015}>\frac{1}{1008.2017}\)
=> \(1+\frac{1}{1008.2015}>1+\frac{1}{1008.2017}\)
=> \(\frac{2015.2016+2}{2015.2016}>\frac{2016.2017+2}{2016.2017}\)
=> \(x>y\)
Ta có:
x = \(\frac{2015.2016+2}{2015.2016}=\frac{2015.2016}{2015.2016}+\frac{2}{2015.2016}=1+\frac{2}{2015.2016}=1+\frac{1}{2015.1008}\)
y = \(\frac{2016.2017+2}{2016.2017}=\frac{2016.2017}{2016.2017}+\frac{2}{2016.2017}=1+\frac{2}{2016.2017}=1+\frac{1}{1008.2017}\)
Do \(\frac{1}{2015.1008}>\frac{1}{1008.2017}\) => \(1+\frac{1}{2015.1008}>1+\frac{1}{1008.2017}\)
=> x > y