NT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
PY
0
22 tháng 7 2016
\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{19.20}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{5}-\frac{1}{20}\)
\(=\frac{4}{20}-\frac{1}{20}=\frac{3}{20}\)
16 tháng 7 2016
\(A=\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+............+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
FK
16 tháng 7 2016
A=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{49.50}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
A=1-\(\frac{1}{50}\)
A=\(\frac{49}{50}\)
TB
27 tháng 1 2018
C=\(\frac{7}{3.4}\)-\(\frac{9}{4.5}\)+\(\frac{11}{5.6}\)+\(\frac{13}{6.7}\)+\(\frac{15}{7.8}\)-\(\frac{17}{8.9}\)+\(\frac{19}{9.10}\)
=\(\frac{1}{3}\)+\(\frac{1}{4}\)-\(\frac{1}{4}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{6}\)-\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)+\(\frac{1}{8}\)-\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)+\(\frac{1}{10}\)
=\(\frac{1}{3}\)+\(\frac{1}{10}\)=\(\frac{13}{30}\)
LV
12 tháng 8 2016
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50
= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50
= 1 - 1/50
= 49/50
ỦNG HỘ NHA
12 tháng 8 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
N
7 tháng 9 2017
\(A=\frac{4}{4.5}+\frac{4}{5.6}+\frac{4}{6.7}+...+\frac{4}{47.48}\)
\(A=4.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+......+\frac{1}{47.48}\right)\)
\(A=4.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{47}-\frac{1}{48}\right)\)
\(A=4.\left(\frac{1}{4}-\frac{1}{48}\right)\)
\(A=4.\frac{11}{48}\)
\(A=\frac{11}{12}\)
7 tháng 6 2016
a) 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/24.25
= 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 4/25
b) 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 -1/101
= 1 - 1/101
= 100/101
c) 3/1.4 + 3/4.7 + ... + 3/2002.2005
= 1 - 1/4 + 1/4 - 1/7 + ... + 1/2002 - 1/2005
= 1 - 1/2005
= 2004/2005
d) 5/2.7 + 5/7.12 + ... + 5/1997.2002
= 1/2 - 1/7 + 1/7 - 1/12 + ... + 1/1997 - 1/2002
= 1/2 - 1/2002
= 500/1001
7 tháng 6 2016
a,A = \(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{24\times25}\)
A\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\)
A\(=\frac{1}{5}-\frac{1}{25}=\frac{5}{25}-\frac{1}{25}=\frac{4}{25}\)
b, B=\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
B= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
B=\(1-\frac{1}{101}=\frac{100}{101}\)
c, \(C=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{2002\times2005}\)
C= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)
C= \(1-\frac{1}{2005}=\frac{2004}{2005}\)
d, D= \(\frac{5}{2\times7}+\frac{5}{7\times12}+...+\frac{5}{1997\times2002}\)
D= \(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{1997}-\frac{1}{2002}\)
D= \(\frac{1}{2}-\frac{1}{2002}=\frac{1001}{2002}-\frac{1}{2002}=\frac{1000}{2002}=\frac{500}{1001}\)
