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= 1 - 1/2 . 1/2 -1/3 . 1/3 - 1/4 ... 1/2009 - 1/2010
= 1 - 1/ 2010
=1/2010
1/1.2+1/2.3+1/3.4+...+1/2009.2010
=1-1/2+1/2-1/3+...+1/2009-1/2010
=1-1/2010
=2009/2010
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}+\frac{1}{2017\cdot2018}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}+\frac{1}{2017}-\frac{1}{2018}\)
\(=2-\frac{1}{2018}\)
\(=\frac{1009}{2018}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\)TỰ RÚT GỌN NHA
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=2-\frac{2007}{2008}\)
\(=\frac{2009}{2008}\)
~Học tốt~
\(I=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2009.2010}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+.....+\left(\frac{1}{2009}-\frac{1}{2009}\right)-\frac{1}{2010}\)
\(I=1-0-0-...-0-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}=\frac{2009}{2010}\)
I = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2009.2010
I = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2009 - 1/2010
I = 1 - 1/2010
I = 2009/2010
Vậy I = 2009/2010
1/1.2+1/2.3+...+1/2009.2010
=1-1/2+1/2-1/3+...+1/2009-1/2010
=1-1/2010
=2009/2010
I=1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010
I=1-1/2010
I=2009/2010
Vậy I=2009/2010
I = 1/1-1/2+1/2-1/3+1/3-1/4+...+1/2009-1/2010
I = 1-1/2010
I = 2009/2010
Chúc bạn học tốt nha
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}\)
\(I=\frac{2009}{2010}\)
Lời giải:
$A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{(n+1)-n}{n(n+1)}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}=\frac{n}{n+1}$
Ta có đpcm.
hí ae toi ms ngủ day