B = 3/10x12+3/12x14+3/14x16+...+3/48x50
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NP
0
PT
2
22 tháng 5 2021
= 3 x ( \(\frac{1}{10x12}+\frac{1}{12x14}+\frac{1}{14x16}+\frac{1}{96x98}\))
= 3 x (\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+......+\frac{1}{96}-\frac{1}{98}\))
= 3 x ( \(\frac{1}{10}-\frac{1}{98}\))
= 3 x \(\frac{22}{245}\)
= \(\frac{66}{245}\)
XO
22 tháng 5 2021
\(\frac{3}{10.12}+\frac{3}{12.14}+\frac{3}{14.16}+...+\frac{3}{96.98}=\frac{3}{2}\left(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{96.98}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{98}\right)\)
\(=\frac{3}{2}\left(\frac{1}{10}-\frac{1}{98}\right)=\frac{3}{2}.\frac{22}{245}=\frac{33}{245}\)
HG
1
C
29 tháng 5 2022
`M=4/(10xx12)+4/(12xx14)+4/(14xx16)+...+4/(96xx98)`
`M=2xx(2/(10xx12)+2/(12xx14)+2/(14xx16)+...+2/(96xx98))`
`M=2xx(1/10-1/12+1/12-1/14+1/14-1/16+...+1/96-1/98)`
`M=2xx(1/10-1/98)`
`M=2xx22/245`
`M=44/245`
25 tháng 5 2018
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
20 tháng 6 2021
Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
7 tháng 5 2020
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
10 tháng 5 2020
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{3}{7}\)
10 tháng 5 2020
\(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\frac{2}{8\times10}+\frac{2}{10\times12}+\frac{2}{12\times14}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}+\frac{1}{10}-\frac{1}{10}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}\)
\(=\frac{1}{2}-\frac{1}{14}\)
\(=\frac{7}{14}-\frac{1}{14}\)
\(=\frac{6}{14}=\frac{3}{7}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 5 2023
\(\dfrac{32}{48}\) \(\times\) \(\dfrac{50}{75}\) = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{2}{3}\) = \(\dfrac{4}{9}\)
⟹2/3B= 2/10.12+3/12.14+...+3/48.50
⟹2/3B=1/10-1/12+1/12-1/14+...+1/48-1/50
⟹2/3B=1/10-1/50
⟹2.3B=4/50
⟹B=4/50:2/3
⟹B=4/50.2/3
⟹B=8/150