Tính giá trị biểu thức:
\(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)
(TRÌNH BÀY CÁCH TÍNH RÕ RÀNG)
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
24 tháng 5 2017
\(=\frac{4+9}{4.9}-\frac{14-9}{9.14}-\frac{14+19}{14.19}+\frac{19+24}{19.24}\)
\(=\frac{4}{4.9}+\frac{9}{4.9}-\frac{14}{9.14}-\frac{9}{9.14}-\frac{14}{14.19}+\frac{19}{14.19}+\frac{19}{19.24}+\frac{24}{19.24}\)
\(\frac{1}{9}+\frac{1}{4}-\frac{1}{9}-\frac{1}{14}-\frac{1}{19}+\frac{1}{14}+\frac{1}{19}+\frac{1}{24}\)
\(=\frac{1}{4}+\frac{1}{24}=\frac{7}{24}\)
AI THẤY ĐÚNG ỦNG HỘ NHA
TM
23 tháng 5 2017
mình nghĩ là đề như vậy:
\(\frac{24}{8.16}-\frac{40}{16.24}+\frac{56}{24.32}-\frac{72}{32.40}=\frac{8+16}{8.16}-\frac{16+24}{16.24}+\frac{24+32}{24.32}-\frac{32+40}{32.40}\)
\(=\frac{8}{8.16}+\frac{16}{8.16}-\frac{16}{16.24}-\frac{24}{16.24}+\frac{24}{24.32}+\frac{32}{24.32}-\frac{32}{32.40}-\frac{40}{32.40}\)
\(=\frac{1}{16}+\frac{1}{8}-\frac{1}{24}-\frac{1}{16}+\frac{1}{32}+\frac{1}{24}-\frac{1}{40}-\frac{1}{32}\)
\(=\frac{1}{8}-\frac{1}{40}=\frac{1}{10}\)
2 tháng 5 2016
A = \(\frac{5}{1.2}\) + \(\frac{5}{2.3}\) +........+\(\frac{5}{99.100}\)
A = 5.(\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +......+\(\frac{1}{99.100}\) )
A = 5. ( \(\frac{1}{1}\) - \(\frac{1}{2}\) +\(\frac{1}{2}-\frac{1}{3}\) +......+\(\frac{1}{99}-\frac{1}{100}\) )
A= 5. (\(1-\frac{1}{100}\))
A= 5.\(\frac{99}{100}\)
A= \(\frac{99}{20}\)
23 tháng 3 2017
B = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+............+ \(\frac{1}{9.10}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)- \(\frac{1}{4}\)+ ...................+\(\frac{1}{9}\)- \(\frac{1}{10}\)
= \(\frac{1}{2}\) - \(\frac{1}{10}\)
= \(\frac{2}{5}\)
6 tháng 6 2015
\(M=\frac{1}{9.10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\right)=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)
PM
1
26 tháng 5 2020
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
Ez :))
PY
0
NT
0
14 tháng 8 2016
\(B=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(B=\frac{1}{3}-\frac{1}{10}\)
\(B=\frac{7}{30}\)
DA
1
3 tháng 7 2015
\(\frac{1}{5.6}\)- \(\frac{1}{6.7}\)- \(\frac{1}{7.8}\) - ... - \(\frac{1}{2004.2005}\)
= \(\frac{1}{5}\)- \(\frac{1}{6}\)+ \(\frac{1}{6}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{8}\)+ ... + \(\frac{1}{2004}\)- \(\frac{1}{2005}\)
=\(\frac{1}{5}\)- \(\frac{1}{2005}\)
= \(\frac{80}{401}\)
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{21}-\frac{1}{22}\)
\(A=\frac{1}{5}-\frac{1}{22}\)
\(A=\frac{17}{110}\)
\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+...+\(\frac{1}{21.22}\)
=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+...+\(\frac{1}{21}\)-\(\frac{1}{22}\)
=\(\frac{1}{5}\)-\(\frac{1}{22}\)
=\(\frac{17}{110}\)