1.4 + 4.7 + 7.9 + .....+ 19.22
Mik cần cực gấp nhé !
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NT
2
29 tháng 4 2023
\(\dfrac{3}{2}\)B= \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)
\(\dfrac{3}{2}\)B= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\) \(\dfrac{3}{2}\)B= \(\dfrac{102}{103}\) \(\)B= \(\dfrac{102}{103}:\dfrac{3}{2}\) B=\(\dfrac{68}{103}\)
LM
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
15 tháng 5 2017
\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)
\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)
\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)
\(B=\frac{102}{3.103}=\frac{34}{103}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
15 tháng 5 2017
3B=3/1.4+3/4.7+3/7.10+...+3/100.103
3B=(4-1)/1.4+(7-4)/4.7+(10-7)/7.10+...+(103-100)/100.103
3B=1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103=1-1/103=102/103
B=102/(3.103)=34/103
NQ
1
10 tháng 3 2017
Ta co:\(\)
\(A=\dfrac{2}{1.4}+\dfrac{2}{4.7}+.....+\dfrac{2}{73.76}\)
\(=>A=2.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+....+\dfrac{1}{73.76}\right)\)
\(=>A=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{73}-\dfrac{1}{76}\right)\)
\(=>A=2.\left(1-\dfrac{1}{76}\right)\)
\(=>A=2.\dfrac{75}{76}=\dfrac{2.75}{2.38}\)
\(=>A=\dfrac{75}{38}\)
Tick cho mk nha
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
NM
Nguyễn Minh Quang
Giáo viên
10 tháng 3 2022
ta nhân 3 cả hai vế, được :
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)
hay
\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)
\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)
29 tháng 6 2017
a) \(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+.....+\frac{5}{27.30}\)
\(=\frac{5}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+........+\frac{1}{27.30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{27}-\frac{1}{30}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{30}\right)\)
\(=\frac{5}{3}.\frac{29}{30}=\frac{29}{36}\)
DL
1 tháng 7 2017
Đặt \(A=\frac{12}{3\cdot5}+\frac{12}{5\cdot7}+\frac{12}{7\cdot9}+....+\frac{12}{97\cdot99}\)
\(2A=\frac{12}{3}-\frac{12}{5}+\frac{12}{5}-\frac{12}{7}+...+\frac{12}{97}-\frac{12}{99}\)
\(2A=\frac{12}{3}-\frac{12}{99}\)
\(A=\frac{128}{33}\cdot\frac{1}{2}=\frac{64}{33}\)
LG
1
PT
6 tháng 4 2016
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
=\(\frac{99}{100}\)
TT
4
20 tháng 8 2016
\(=\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{1}{3}.\left(\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(=\frac{1}{3}.\left(1-\frac{1}{100}\right)=\frac{x}{2}\)
\(\frac{1}{3}.\frac{99}{100}=\frac{x}{2}\)
\(\frac{99}{300}=\frac{x}{2}\)
\(x\)ko thỏa mãn
LM
20 tháng 8 2016
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.9}+......+\frac{1}{97.100}=\frac{x}{2}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{97}-\frac{1}{100}\right)=\frac{x}{2}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{100}\right)=\frac{x}{2}\Rightarrow\frac{1}{3}.\frac{99}{100}=\frac{x}{2}\Rightarrow\frac{33}{100}=\frac{x}{2}\Rightarrow\frac{33}{100}=\frac{50x}{100}\Rightarrow33=50x\Rightarrow x=\frac{33}{50}\)
đề ???
Đăt A = 1.4 + 4.7 + 7.10 + ....+ 19.22
=> 9A = 1.4.9 + 4.7.9 + 7.10.9 + ...+ 19.22.9
9A = 1.4.(7+2) + 4.7.(10-1) + 7.10.(13-4) + ...+ 19.22.(25-16)
9A = 1.4.7 + 4.2 + 4.7.10 - 1.4.7 + 7.10.13 - 4.7.10 + ....+ 19.22.25 - 16.19.22
9A = 4.2 + 19.22.25
A = 1 162