E=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.........+\frac{1}{86.87.88}\)
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4 tháng 8 2018
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
= \(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{20}\)
=\(\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)
NM
22 tháng 3 2019
Xét ct trước :D
\(\frac{2}{\left[\left(n-1\right)n\left(n+1\right)\right]}=\frac{1}{\left[\left(n-1\right)n\right]}-\frac{1}{\left[n\left(n+1\right)\right]}\)
Sau khi xét ct rồi thì /Bùm/ Ta được:
\(2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{1.2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
\(\Rightarrow M=\frac{65}{264}\)
Ok rồi nhé :)
15 tháng 4 2019
A = 5/20.22 + 5/22.24+...+5/79.81
A = 5/2 . (2/20.22 + 2/22.24 + ... + 2/79.81)
A = 5/2 . (1/20 - 1/22 + 1/22 - 1/24 + ... + 1/79 - 1/81)
A = 5/2 . (1/20 - 1/81)
A = 5/2 . 61/1620
A = 61/648
15 tháng 4 2019
B = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.29
2B = 2/1.2.3 + 2/2.3.4 + ... + 2/18.19.20
\(\Rightarrow\)B = 1/1.2 + 1/2.3 + ... + 1/19.20
\(\Rightarrow\)B = 1/1.2 - 1/19.20
B = 1/2 - 1/380
B = 189/380
HP
21 tháng 2 2016
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{22}{45}\)
\(\Leftrightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\Leftrightarrow\frac{22}{45}.x=\frac{44}{45}\Leftrightarrow x=2\)
Vậy x=2
6 tháng 3 2019
\(D=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{10\cdot11\cdot12}\)
\(D=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{10\cdot11\cdot12}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\right)\)
\(D=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{132}\right)=...\)
IA
6 tháng 3 2019
\(D=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
\(D=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{10.11.12}\right).\frac{1}{2}\)
\(D=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right).\frac{1}{2}\)
\(D=\left(\frac{1}{1.2}-\frac{1}{11.12}\right).\frac{1}{2}\)
\(D=\frac{65}{132}.\frac{1}{2}\)
\(D=\frac{65}{264}\)
D
12 tháng 10 2019
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{8\cdot9\cdot10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}\cdot\frac{22}{45}=\frac{22}{45}\)
\(\Rightarrow\frac{x}{2}=1\)
\(\Rightarrow x=2\)
Từ giả thiết suy ra:
2E=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+..+\frac{2}{86.87.88}\)
2E=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{86.87}+\frac{1}{87.88}\)
2E=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{86}-\frac{1}{87}+\frac{1}{87}-\frac{1}{88}\)
2E=\(1-\frac{1}{88}\)
2E=\(\frac{87}{88}\)
E=\(\frac{87}{176}\)
Vậy E=\(\frac{87}{176}\)