1.1/2x4x6+1/4x6x8+1/6x8x10+....+1/96x98x180
2.(1-3/4) x (1-3/7) x (1-3/10) x...x(1x3/97)x(1-3/100)
Ae giúp tôi, mai nộp r
K
Khách
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DP
2 tháng 8 2017
\(\frac{1}{2.4.6}+\frac{1}{4.6.8}+\frac{1}{6.8.10}+..+\frac{1}{50.52.54}\)
\(=\frac{1}{4}.\left(\frac{1}{2.4}-\frac{1}{4.6}+\frac{1}{4.6}-\frac{1}{6.8}+....+\frac{1}{50.52}-\frac{1}{52.54}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{2.4}-\frac{1}{52.54}\right)\)
\(=\frac{1}{4}.\frac{175}{1404}=\frac{175}{5616}\)
LB
1
NM
1
NN
22 tháng 10 2015
Đặt biểu thức là A
\(4xA=\frac{4}{2x4x6}+\frac{4}{4x6x8}+\frac{4}{6x8x10}+\frac{4}{8x10x12}+...+\frac{4}{94x96x98}+\frac{4}{96x98x100}\)
\(4xA=\frac{6-2}{2x4x6}+\frac{8-4}{4x6x8}+\frac{10-6}{6x8x10}+\frac{12-8}{8x10x12}+...+\frac{98-94}{94x96x98}+\frac{100-96}{96x98x100}\)
\(4xA=\frac{1}{2x4}-\frac{1}{4x6}+\frac{1}{4x6}-\frac{1}{6x8}+\frac{1}{6x8}-\frac{1}{8x10}+...+\frac{1}{94x96}-\frac{1}{96x98}+\frac{1}{96x98}-\frac{1}{98x100}\)
\(4xA=\frac{1}{2x4}-\frac{1}{98x100}=\frac{49x50-1}{98x100}\Rightarrow A=\frac{49x50-1}{4x98x100}\)
17 tháng 6 2021
Giải:
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1.2.3.4}{2.3.4.5}\)
\(=\dfrac{1}{5}\)
b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Chúc bạn học tốt!
KM
0
2
21 tháng 12 2022
`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
17 tháng 6 2021
Giải:
\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
DN
4
T
28 tháng 5 2019
#)Giải :
\(\left(1-\frac{3}{4}\right)x\left(1-\frac{3}{7}\right)x\left(1-\frac{3}{10}\right)x\left(1-\frac{1}{13}\right)x...x\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}x\frac{4}{7}x\frac{7}{10}x...x\frac{94}{97}x\frac{97}{100}\)
\(=\frac{1x4x7x...x94x100}{4x7x10x...x97x100}\)
\(=\frac{1}{100}\)
#~Will~be~Pens~#
KN
28 tháng 5 2019
\(\left(1-\frac{3}{4}\right)\left(1-\frac{3}{7}\right)\left(1-\frac{3}{10}\right)\left(1-\frac{1}{13}\right)...\left(1-\frac{1}{97}\right)\left(1-\frac{3}{100}\right)\)
\(=\frac{1}{4}.\frac{4}{7}.\frac{7}{10}.\frac{10}{13}...\frac{94}{97}.\frac{97}{100}\)
\(=\frac{1}{100}\)
