2/5.8+2/8.11+2/11.14+....+2/95.98
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Những câu hỏi liên quan
HN
1
28 tháng 4 2015
A=2/2.5+2/5.8+2/8.11+...+2/95.98
=2/3.(3/2.5+3/5.8+3/8.11+...+3/95.98)
=2/3.(1/2-1/5+1/5-1/8+1/8-1/11+...+1/95-1/98)
=2/3.(1/2-1/98)
=2/3.24/49
=16/49
VẬY A=16/49
J
2
NH
20 tháng 3 2023
\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)
20 tháng 3 2023
A=21.3+23.5+...+297.99�=21.3+23.5+...+297.99
A=11−13+13−15+...+197−199�=11−13+13−15+...+197−199
A=11−199�=11−199
A=9899
=
21−51+51−71+....+951−981
=12−198=21−981tự làm tiếp nha ( giống câu a)
NY
4
10 tháng 5 2019
\(A=\frac{2}{2.5}+\frac{2}{5.8}+...+\frac{2}{95.98}\)
\(A=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{95.98}\right)\)
\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{2}{3}.\frac{24}{49}\)
\(A=\frac{16}{49}\)
KN
11 tháng 5 2019
\(A=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{95.98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{95.98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{1}{2}-\frac{1}{98}\)
\(\Leftrightarrow\frac{3}{2}A=\frac{48}{98}=\frac{24}{49}\)
\(\Leftrightarrow A=\frac{24}{49}\div\frac{3}{2}\)
\(\Leftrightarrow A=\frac{48}{147}\)
20 tháng 8 2023
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
1 tháng 4 2023
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
LT
6 tháng 3 2020
\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)
=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{17}\right)\)
=\(\frac{45}{34}\)
\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
13 tháng 7 2021
=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)
=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)
=3(1/2-1/17)
=45/34
cô Nhung ơi k đúng cho con đi cô pls
TH
1
7 tháng 8 2017
Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{32.35}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{32}-\frac{1}{35}\)
\(A=\frac{1}{5}-\frac{1}{35}=\frac{6}{35}\)
\(\Rightarrow x+\frac{6}{35}=-\frac{2}{7}\Rightarrow x=-\frac{2}{7}-\frac{6}{35}=-\frac{16}{35}\)
18 tháng 9 2020
\(3x-\frac{15}{5\cdot8}-\frac{15}{8\cdot11}-\frac{15}{11\cdot14}-...-\frac{15}{47\cdot50}=2\frac{1}{10}\)
<=> \(3x-5\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{47\cdot50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{47}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\left(\frac{1}{5}-\frac{1}{50}\right)=\frac{21}{10}\)
<=> \(3x-5\cdot\frac{9}{50}=\frac{21}{10}\)
<=> \(3x-\frac{9}{10}=\frac{21}{10}\)
<=> \(3x=3\)
<=> \(x=1\)
AM
20 tháng 6 2015
\(H=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{197.200}\right)=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{197}-\frac{1}{200}\right)=3\cdot\left(\frac{1}{2}-\frac{1}{200}\right)==\frac{297}{200}\)
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