a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2008.2009}\)
b) \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{94.97}\)
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Những câu hỏi liên quan
24 tháng 3 2019
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
24 tháng 3 2019
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
HH
15 tháng 6 2018
Giải:
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}\)
\(=\dfrac{2008}{2009}\)
c) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{4}{7.10}+...+\dfrac{3}{94.97}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=\dfrac{1}{1}-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
Vậy ...
Các câu sau tương tự
16 tháng 6 2018
b, \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2008.1009}\)
\(=\)\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(=\dfrac{1}{1}-\dfrac{1}{2009}=\dfrac{2009}{2009}-\dfrac{1}{2009}=\dfrac{2008}{2009}\)
DP
19 tháng 7 2017
\(S=\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(S=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
\(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(T=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+....+\frac{43-40}{40.43}+\frac{46-43}{43.46}\)
\(T=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(T=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)
18 tháng 7 2017
1) Ta có : \(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Vậy T = \(=\frac{99}{100}\)
2) Ta có : \(T=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}=\frac{45}{46}\)
Vậy T = \(\frac{45}{46}\)
27 tháng 8 2017
a) = 1-1/2+1/2-1/3+1/3-1/4
= 1-1/4=3/4
b)=1-1/2+1/2-1/3+1/3-1/4+...+1/2016-1/2017+1/2017-1/2018
=1-1/2018=2017/2018
c)=1/2-1/5+1/5-1/8+1/8-1/11+1/2009-1/2012+1/2012-1/2015
= 1/2-1/2015=2015/4030-2/4030=2013/4030
Q
27 tháng 8 2017
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017-2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
c) \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2012.2015}\)
\(=3\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{2012.2015}\right)\)
\(\Leftrightarrow\frac{3}{2}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2012}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}.\frac{2013}{4030}\)
\(=\frac{6039}{8060}\)
T
25 tháng 3 2018
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
9 tháng 4 2015
a)\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
=\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{91}-\frac{1}{94}+\frac{1}{94}-\frac{1}{97}\)(giản ước các phân số giống nhau)
=\(\frac{1}{1}-\frac{1}{97}\)
=\(\frac{96}{97}\)
9 tháng 4 2015
a) gọi \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.11}+...+\frac{2}{94.97}\)
\(\Rightarrow\frac{3}{2}A=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}\)
\(\frac{3}{2}A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)(rút gọn các phân số giống nhau)
\(\frac{3}{2}A=\frac{1}{1}-\frac{1}{97}\)
\(\frac{3}{2}A=\frac{96}{97}\left(1\right)\)
từ \(\left(1\right)\Leftrightarrow A=\frac{96}{97}\div\frac{3}{2}=\frac{64}{97}\)
b)\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{2011}\right)\)
\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}......\frac{2010}{2011}\)
\(=\frac{6.7.8.9.....2010}{7.8.9......2011}\)(rút gọn các số giống nhau)
\(=\frac{6}{2011}\)
22 tháng 4 2015
s=(1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103)+(1/103-1/104+1/104-1/105+1/105-1/106+1/106-1/107)
=(1-1/103)+(1/103-1/107)
=1 - 1/107
=106/107
17 tháng 3 2018
a, 1/1.2+1/1.3+...+1/99.100
= 1-1/2+1/2-1/3+1/3+...+1/99-1/100
=1-1/100
=99/100
21 tháng 3 2018
c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)
1/2-1/94 ban tu tinh nhe
d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)
1-1/94 ban tu tinh nhe
tk nha
TL
21 tháng 3 2018
a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
c,d dễ bn tách ra rồi trừ đi
a)=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2009}\)
\(=1-\frac{1}{2009}\)
\(=\frac{2008}{2009}\)
b) =\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
=\(\frac{96}{97}\)
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2008.2009}\) \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2008}-\frac{1}{2009}\)
= 1 - 1/2009
= 2008/2009
b) 3/1.4 + 3/4.7 + 3/7.10 + .... + 3/94.97
= 1- 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/94 - 1/97
= 1 - 1/97
= 96/97