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PA
5
23 tháng 8 2019
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\right).\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{97}\right)\)
\(=\frac{1}{3}.\frac{96}{97}\)
\(=\frac{32}{97}\)
học tốt
23 tháng 8 2019
3A = 3(1/1.4 + 1/4.7 + 1/7.10 + ...... + 1/94.97)
3A=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - ........ - 1/97
3A = 1-1/97
3A = 96/97
A = 32/97
Oke nha bạn
VN
2
26 tháng 4 2018
\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
mà \(\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)
vậy..................
26 tháng 4 2018
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
\(\Rightarrow\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
NM
1
3 tháng 5 2019
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)
\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)
Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)
Vậy:.............................<1
3 tháng 5 2020
\(S_1=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{48\cdot49}+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
\(S_2=\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+....+\frac{1}{94\cdot97}+\frac{1}{97\cdot100}\)
\(3S_2=\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+....+\frac{3}{94\cdot97}+\frac{3}{97\cdot100}\)
\(=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)
=> \(S_2=\frac{6}{25}:3=\frac{2}{25}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 3
a; \(\dfrac{-1}{n}\) - \(\dfrac{1}{n+a}\)
= \(\dfrac{-n-a-n}{n.\left(n+a\right)}\)
= \(\dfrac{-2n-a}{n.\left(n+a\right)}\)
b; \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ....+ \(\dfrac{1}{2007.2008}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{2008}\)
= \(\dfrac{2007}{2008}\)
c; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
= \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
20 tháng 8 2023
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
N
14 tháng 7 2015
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) =\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
=\(\frac{96}{97}\)
14 tháng 7 2015
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
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)
\(=\frac{6}{2011}\)
C
2 tháng 3 2023
`3/1.4+3/4.7+3/7.10+...+3/94.97`
`=1/1-1/4+1/4-1/7+1/7-1/10+...+1/94-1/97`
`=1-1/97`
`=96/97`
2 tháng 3 2023
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)