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29 tháng 3 2018
Ta có :
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{65.68}\)
\(=\)\(4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{65.68}\right)\)
\(=\)\(4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{65}-\frac{1}{68}\right)\)
\(=\)\(4\left(\frac{1}{2}-\frac{1}{68}\right)\)
\(=\)\(2-\frac{1}{17}\)
\(=\)\(\frac{35}{17}\)
Vậy \(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{65.68}=\frac{35}{17}\)
Chúc bạn học tốt ~
19 tháng 7 2018
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+...+\frac{12}{29.32}\)
\(=4.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=4.\frac{15}{32}\)
\(=\frac{15}{8}\)
_Chúc bạn học tốt_
KT
19 tháng 7 2018
\(\frac{12}{2.5}+\frac{12}{5.8}+\frac{12}{8.11}+....+\frac{12}{29.32}\)
\(=4\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{29.32}\right)\)
\(=4\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)
\(=4\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(=4.\frac{15}{32}=\frac{15}{8}\)
24 tháng 7 2021
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
MV
8 tháng 1 2018
\(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+\dfrac{4}{8\cdot11}+...+\dfrac{4}{65\cdot68}\\ =\dfrac{4}{3}\cdot\dfrac{3}{2\cdot5}+\dfrac{4}{3}\cdot\dfrac{3}{5\cdot8}+\dfrac{4}{3}\cdot\dfrac{3}{8\cdot11}+...+\dfrac{4}{3}\cdot\dfrac{3}{65\cdot68}\\ =\dfrac{4}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{65\cdot68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\dfrac{33}{68}\\ =\dfrac{11}{17}\)
VN
Võ Ngọc Phương
VIP
25 tháng 8 2023
Sửa đề:
\(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(A=4.\left(\dfrac{34}{68}-\dfrac{1}{68}\right)\)
\(A=4.\dfrac{33}{68}\)
\(A=\dfrac{33}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\)+ \(\dfrac{4}{8.11}\)+...+ \(\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}\).( \(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\)+ \(\dfrac{3}{8.11}\)+....+ \(\dfrac{3}{65.68}\))
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\)+...+ \(\dfrac{1}{65}\)- \(\dfrac{1}{68}\)
A = \(\dfrac{4}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
A = \(\dfrac{4}{3}\). \(\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
NT
1
7 tháng 7 2015
9D=2.5.9+5.8.9+8.11.9+11.14.9+...+62.65.9+65.68.9
9D=2.5.9+5.8.(11-2)+8.11.(14-5)+11.14.(17-8)+...+62.65.(68-59)+65.68.(71-62)
9D=2.5.9+5.8.11-2.5.8+8.11.14-5.8.11+11.14.17-8.11.14+...+62.65.68-59.62.65+65.68.71-62.65.67
Rút gọn các phần giống nhau còn
9D=2.5.9-2.5.8+65.68.71
9D=10+65.68.71
9D=10+313820
9D=313830
D=313830:9=34870
Vậy D=34870
VM
3
28 tháng 4 2018
A = \(\dfrac{4}{2.5}\) + \(\dfrac{4}{5.8}\) + \(\dfrac{4}{8.11}\) + ... + \(\dfrac{4}{65.68}\)
7A = \(\dfrac{4.3}{2.5}\) + \(\dfrac{4.3}{5.8}\) + \(\dfrac{4.3}{8.11}\) + ... + \(\dfrac{4.3}{65.68}\)
7A = 4 (\(\dfrac{3}{2.5}\) + \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\) + ... + \(\dfrac{3}{65.68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + ... + \(\dfrac{1}{65}\) - \(\dfrac{1}{68}\))
7A = 4 (\(\dfrac{1}{2}\) - \(\dfrac{1}{68}\))
7A = 4 . \(\dfrac{33}{68}\) = \(\dfrac{33}{17}\)
A = \(\dfrac{33}{17}\) : 7
=> A = \(\dfrac{33}{119}\)
TN
28 tháng 4 2018
Ta có: \(A=\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{5-2}{2.5}+\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+...+\dfrac{68-65}{65.68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)=\dfrac{4}{3}.\dfrac{33}{68}=\dfrac{11}{17}\)
NT
2
14 tháng 2 2016
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{65}-\frac{1}{68}\right)\)
\(\Rightarrow A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{68}\right)=\frac{1}{2}\left(\frac{34}{68}-\frac{1}{68}\right)=\frac{1}{2}.\frac{33}{68}=\frac{33}{136}\)
PQ
6
9 tháng 8 2015
S = 4/2.5 + 4/5.8 + 4/8.11 + ... + 4/65.48
S = 4/3 . ( 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/65.68 )
S = 4/3 . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/65 - 1/68 )
S = 4/3 . ( 1/2 - 1/68 )
S = 4/3 . 33/68
S = 11/17
\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+.......+\dfrac{12}{65.68}\)
\(=4\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+......+\dfrac{3}{65.68}\right)\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+.......+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=4\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=4.\dfrac{33}{68}=\dfrac{33}{17}\)
Dễ quá! Vì mình là một CTV bên Học toán với OnlineMath nên bài này easy!! :")))
\(\dfrac{12}{2.5}+\dfrac{12}{5.8}+\dfrac{12}{8.11}+...+\dfrac{12}{65.68}\)
\(\Leftrightarrow4\left(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{65.68}\right)\)
\(\Leftrightarrow4\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=4\left(1-\dfrac{1}{68}\right)=4.\dfrac{67}{68}=\dfrac{67}{17}\)