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LN
8
NH
16 tháng 5 2019
\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(S=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2020}\right)\)
Tự tính
17 tháng 5 2019
S=1/2.4+1/4.6+1/6.8+...+1/2018.2020
S=1/2.(2/2.4+2/4.6+2/6.8+...+2/2018.2020)
S=1/2.(1-1/4+1/4-1/6+1/6-1/8+...+1/2018-1/2020)
S=1/2.(1-1/2020)
S=1/2.(2020/2020-1/2020)
S=1/2.2019/2020
S=2019/4040
D
4
NH
30 tháng 7 2019
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(=\frac{1}{2}-\frac{1}{2020}=\frac{1009}{2020}\)
30 tháng 7 2019
\(A=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2018.2020}\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+....+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)=\frac{1}{2}.\frac{1009}{2020}\)
\(\Leftrightarrow A=\frac{1009}{4040}\)
Vậy : \(A=\frac{1009}{4040}\)
NT
1
29 tháng 4 2021
Sửa đề: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
Ta có: \(\dfrac{4}{2\cdot4}+\dfrac{4}{4\cdot6}+\dfrac{4}{6\cdot8}+...+\dfrac{4}{2018\cdot2020}+\dfrac{4}{2020\cdot2022}\)
\(=2\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2018\cdot2020}+\dfrac{2}{2020\cdot2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2018}-\dfrac{1}{2020}+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=2\cdot\dfrac{505}{1011}\)
\(=\dfrac{1010}{1011}\)
14 tháng 6 2019
a,A =\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{199}-\frac{1}{200}\)
= 1-\(\frac{1}{200}\)
=\(\frac{199}{200}\)
14 tháng 6 2019
b, B=\(\frac{3}{2.4}+\frac{3}{4.6}+\frac{3}{6.8}+...+\frac{3}{2018.2020}\)
=3.(\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+..+\frac{1}{2018.2020}\))
=3(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2018}-\frac{1}{2020}\))
= 3.(\(\frac{1}{2}-\frac{1}{2020}\))
=\(\frac{6057}{2020}\)
1 tháng 8 2019
a, \(A=\frac{6}{10.11}+\frac{6}{11.12}+\frac{6}{12.13}+...+\frac{6}{69.70}\)
\(A=\frac{6}{10}-\frac{6}{11}+\frac{6}{11}-\frac{6}{12}+\frac{6}{12}-\frac{6}{13}+...+\frac{6}{69}-\frac{6}{70}\)
\(A=\frac{6}{10}-\frac{6}{70}\)
\(A=\frac{18}{35}\)
b, \(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(B=\frac{4}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(B=2.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(B=2.\frac{1009}{2020}\)
\(B=\frac{1009}{1010}\)
Chúc bạn học tốt 
6 tháng 10 2019
Hơi thắc mắc câu B cậu oi!!!Gỉai thích cho mk vs ạ!!Thanks
SA
17 tháng 3 2019
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
SA
17 tháng 3 2019
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
HB
1
HG
15 tháng 12 2016
A=1/2.4+1/4.6+........+1/100.102
A=1/2-1/4+1/4-1/6+.......+1/100-1/102
A=1/2-1/102
A=51/102-1/102
A=50/102
A=25/51
TN
22 tháng 3 2016
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{18}-\frac{1}{20}\)
\(A=\frac{1}{2}-\frac{1}{20}\)
\(A=\frac{10}{20}-\frac{1}{20}\)
\(A=\frac{9}{20}\)
\(A=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2018.2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2018}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2020}\right)\)
\(A=\frac{1}{2}.\frac{1009}{2020}\)
\(A=\frac{1009}{4040}\)
A=1/2.4+1/4.6+1/6.8+...+1/2018.2020
=1/2(1/2-1/4+1/4-1/6+...+1/2018-1/2020)
=1/2(1/2-1/2020)
=1/2.1009/2020
=1009/4040