K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
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
30 tháng 6 2017
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{94.97}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(=1-\dfrac{1}{97}\)
\(=\dfrac{96}{97}\)
QT
Quoc Tran Anh Le
Giáo viên
15 tháng 12 2017
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\right)\)
\(=3\left(1-\dfrac{1}{97}\right)\)
\(=3.\dfrac{96}{97}=\dfrac{288}{97}\)
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; \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
= \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
= \(\dfrac{1}{1}\) - \(\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
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
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
DT
3
18 tháng 5 2020
Đặt 2/3 ra ngoài trong ngoặc còn :
1-1/4+1/4-1/7+...-1/97=96/97
Lấy 2/3 nhân với 96/97 sẽ ra đáp án nhé
K
1
16 tháng 3 2016
3/1.4+3/4.7+....+3/97.100
= 1-1/4+1/4-1/7+....+1/97-1/100
=1-1/100
=99/100
O
25 tháng 3 2022
\(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+\dfrac{3}{13\cdot16}\)
\(=\dfrac{3\cdot1}{1\cdot4}+\dfrac{3\cdot1}{4\cdot7}+\dfrac{3\cdot1}{7\cdot10}+\dfrac{3\cdot1}{10\cdot13}+\dfrac{3\cdot3}{13\cdot16}\)
\(=3\cdot\left(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(=3\cdot\left(1-\dfrac{1}{16}\right)\)
\(=3\cdot\left(\dfrac{16}{16}-\dfrac{1}{16}\right)\)
\(=3\cdot\dfrac{15}{16}\)
\(=\dfrac{45}{16}\)
`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`
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\\ =1-\dfrac{1}{97}=\dfrac{96}{97}\)