(1/2.5 +1/5.8 + 1/811+...+1/17.20)
K
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Những câu hỏi liên quan
NP
1
NP
20 tháng 3 2016
1938.(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938.3/3(1/2.5+1/5.8+1/8.11+...+1/17.20)
= 1938/3.(3/2.5+3/5.8+3/8.11+...+3/17.20)
= 1938/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11+...+1/17-1/20)
= 1938/3.(1/2-1/20)
= 1938/3.9/20
= 2907/10
NV
1
CX
12 tháng 5 2016
S = 1/2.5 +1/5.8 +1/8.11+1/11.14+1/14.17+1/17.20
S=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
S=1/3.(1/2-1/20)
S=1/3.(10/20-1/20)
S=1/3.9/20
S= 3/20
k nha
28 tháng 3 2017
\(\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right]\)
\(\frac{1}{3}\left[\frac{1}{2}-\frac{1}{20}\right]=\frac{1}{3}.\frac{9}{20}=\frac{3}{20}\)
mk đầu tiên đó
TM
12 tháng 5 2017
A=...
<=>\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{1}{17.20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{20}\right)\)
<=>\(A=\frac{1}{6}-\frac{1}{60}< \frac{1}{6}< 1\)
KC
7
23 tháng 6 2018
\(S=\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{17.20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}\)
\(\Rightarrow3S=\frac{9}{20}\)
\(\Rightarrow S=\frac{3}{20}\)
23 tháng 6 2018
\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{17\cdot20}\)
\(S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(S=\frac{1}{2}-\frac{1}{20}\)
\(S=\frac{9}{20}\)
16 tháng 5 2016
S = 1/3 . (1/2 - 1/5 + 1/5 - 1/8 + ... + 1/17 - 1/20)
= 1/3 . (1/2 - 1/20)
= 1/3 . 9/20
= 3/20
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
16 tháng 5 2016
\(3S=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{20-17}{17.20}\)
\(3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
\(S=\frac{9}{20}:3=\frac{3}{20}\)
CB
4
MV
7 tháng 5 2017
\(S=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{17\cdot20}\\ =\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{17\cdot20}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\\ =\dfrac{1}{3}\cdot\dfrac{9}{20}\\ =\dfrac{3}{20}\)
7 tháng 5 2017
Giải:
Ta có:
\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{17.20}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)
\(=\dfrac{1}{3}.\dfrac{9}{20}=\dfrac{3}{20}\)
Vậy \(S=\dfrac{3}{20}\)
S
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{17.20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{17}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\left(\dfrac{1}{2}-\dfrac{1}{20}\right).x=\dfrac{45}{23}\)
\(\dfrac{9}{20}\cdot x=\dfrac{45}{23}\)
\(x=\dfrac{45}{23}:\dfrac{9}{20}\)
\(x=\dfrac{100}{23}\)
TA
7 tháng 4 2023
\(\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{1}{17.20}\right)x=\dfrac{45}{23}\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)\(\Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{20}\right)x=\dfrac{45}{23}\)
\(\Rightarrow\dfrac{9}{20}.x=\dfrac{45}{23}\)
\(\Rightarrow x=\dfrac{100}{23}\)
đặt A=1/2.5 +1/5.8 + 1/811+...+1/17.20
\(3A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(3A=\frac{1}{2}-\frac{1}{20}\)
\(A=\frac{9}{20}:3\)
\(A=\frac{3}{20}\)