CG
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Khách
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12 tháng 4 2023
b: 6B=2*4*6+4*6*6+6*8*6+...+46*48*6+48*50*6
=2*4*6-2*4*6+4*6*8-4*6*8+...-44*46*48+46*48*50-46*48*50+48*50*52
=48*50*52
=>B=20800
d: 9D=1*4*9+4*7*9+...+46*49*9
=1*4*2+1*4*7-1*4*7+1*7*10-1*7*10+...+46*49*52-46*49*43
=1*2*4+46*49*52
=117216
=>D=13024
a: 
NN
2
S
7 tháng 10 2016
Đặt \(A=\frac{1}{1.3}+\frac{1}{2.4}+...+\frac{1}{8.10}\)
\(2A=\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\)
\(2A=1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{10}\)
\(2A=1-\frac{1}{10}\)
\(2A=\frac{9}{10}\)
\(A=\frac{9}{10}:2=\frac{9}{20}\)
7 tháng 10 2016
=\(\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{8.10}\right)\)
=\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}...+\frac{1}{8}-\frac{1}{10}\right)\)
( chắc chắn có số trái dấu ở phía sau, nên còn lại như sau)
=\(\frac{1}{2}\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)
DH
0
25 tháng 5 2022
\(A=\dfrac{1}{2}.\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(\dfrac{1}{2015.2017}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right)....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1}.\dfrac{2}{3}\right).\left(\dfrac{3}{2}.\dfrac{3}{4}\right).\left(\dfrac{4}{3}.\dfrac{4}{5}\right).....\left(\dfrac{2016}{2015}.\dfrac{2016}{2017}\right)\)
\(=\dfrac{2016}{2017}\)
MH
17 tháng 10 2023
\(A=\dfrac{1}{2}\left(\dfrac{2.2}{1.3}\right).\left(\dfrac{3.3}{2.4}\right)...\left(\dfrac{2020.2020}{2019.2021}\right)\)
\(=\dfrac{1.2.2.3.3...2020.2020}{1.2.2.3.3.4.4...2019.2021}\)
\(=\dfrac{1}{2021}\)
17 tháng 10 2023
\(A=\dfrac{1}{2}\cdot\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\left(1+\dfrac{1}{3\cdot5}\right)...\left(1+\dfrac{1}{2019\cdot2021}\right)\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\left(1+\dfrac{1}{4^2-1}\right)...\left(1+\dfrac{1}{2020^2-1}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\cdot\left(3+1\right)}...\left(\dfrac{2020^2}{\left(2020-1\right)\cdot\left(2020+1\right)}\right)\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{2}{3}\cdot\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot...\cdot\dfrac{2020}{2019}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2020}{2021}\)
\(A=\dfrac{1}{2}\cdot2020\cdot\dfrac{2}{2021}=\dfrac{2020}{2021}\)
24 tháng 10 2016
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 9 2023
P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))
P = \(\dfrac{1.3+1}{1.3}\). \(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)
P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)
P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)
P = \(\dfrac{2.2021}{2022}\)
P = \(\dfrac{2021}{1011}\)
NL
2
10 tháng 9 2016
Bạn giải chỉ tiết ra đi. Nêu bạn giải chi tiết mình tích đúng cho
9 tháng 4 2017
2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)
2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)
2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)
2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)
2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)
2A = \(\frac{4032}{2017}\)
A = \(\frac{4032}{2017}\)\(:2\)
A = \(\frac{2016}{2017}\)