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NL
2
24 tháng 2 2022
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)=\dfrac{4}{3}\cdot\dfrac{303}{610}=\dfrac{202}{305}\)
24 tháng 2 2022
\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{302.305}\)
\(=4\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{302.305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{302.305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{305}\right)\)
\(=\dfrac{4}{3}.\dfrac{303}{610}\\ =\dfrac{202}{305}\)
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
ST
19 tháng 4 2021
Ta có : B = 4 / 2. 5 + 4 / 5 . 8 + 4 / 8 . 11 + 4 / 11 . 14
= 4 ( 1 / 2 . 5 + 1 / 5 . 8 + 1 / 8 . 11 + 1 / 11 . 14 )
= 4 . [ 1/3 ( 1 /2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/ 11 + 1/11 - 1/14 )]
= 4 / 3 ( 1/ 2 - 1/14 )
= 4/3 . 3/7
= 4 / 7
Vậy B = 4 / 7
TM
1
CX
14 tháng 5 2016
M = 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20
M= 4/3 . (1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
M= 4/3 . (1/2 - 1/20)
M= 4/3 . (10/20 - 1/20)
M= 4/3 . 9/20
M= 3/5
k nha
24 tháng 7 2021
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
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 = \(\dfrac{11}{17}\)
MV
8 tháng 1 2018
\(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+\dfrac{4}{8\cdot11}+...+\dfrac{4}{65\cdot68}\\ =\dfrac{4}{3}\cdot\dfrac{3}{2\cdot5}+\dfrac{4}{3}\cdot\dfrac{3}{5\cdot8}+\dfrac{4}{3}\cdot\dfrac{3}{8\cdot11}+...+\dfrac{4}{3}\cdot\dfrac{3}{65\cdot68}\\ =\dfrac{4}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{65\cdot68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\\ =\dfrac{4}{3}\cdot\dfrac{33}{68}\\ =\dfrac{11}{17}\)
VH
1
\(=\frac{3}{4}\cdot\left(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{62.65}\right)\)
\(=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{62.65}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{62}-\frac{1}{65}\)
\(=\frac{1}{2}-\frac{1}{65}\)
\(=\frac{63}{130}\)
Đặt A=4/2.5+4/5.8+4/8.11+...+4/62.65.Ta có A=4.(1/2.5+1/5.8+1/8.11+...1/62.65)=4/3.(3/2.5+3/5.8+3/8.11+...+3/62.65) =4/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+3/62-3/65)=4/3.(1/2-1/65)=4/3.63/130=42/56 Vậy A=42/56