\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}\)
Mấy bạn giúp mik nha
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
B=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2016.2017}\)
C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2016.2018}\)
D=\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
E=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}\cdot...\cdot\frac{899}{900}\)
F=1.2+2.3+3.4+...+99.100
MẤY BN NÀO BIẾT THÌ GIẢI JUP MK NHA!
Lâm đi là: 35 phút +2 giờ 20phút =2 giờ 55 phút
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(A=1-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2016.2018}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2016.2018}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(2B=1-\frac{1}{2018}\)
\(B=\frac{2017}{\frac{2018}{2}}\)
Rút gọn:
a/ \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2000}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{1998.1999.2000}\)
c/ \(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2006.2008}\)
a)A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2009.2010}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2009}-\frac{1}{2010}\)
A=1-\(\frac{1}{2010}\)=\(\frac{2009}{2010}\)
c)C=\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+......+\frac{1}{2006.2008}\)
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+..+\frac{1}{2006}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).(\(\frac{1}{2}-\frac{1}{2008}\))
C=\(\frac{1}{2}\).\(\frac{1003}{2008}\)=\(\frac{1003}{4016}\)
Câu b mình chưa nghĩ ra
Chúc bạn học tốt!
a) A = \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + ...+ \(\frac{1}{2009.2000}\)
= 1 - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ... + \(\frac{1}{2009}\) - \(\frac{1}{2000}\)
= 1 - \(\frac{1}{2000}\) = \(\frac{1999}{2000}\)
b) B = \(\frac{1}{1.2.3}\) + \(\frac{1}{2.3.4}\) + \(\frac{1}{3.4.5}\) + ... + \(\frac{1}{1998.1999.2000}\)
= \(\frac{1}{2}\) ( \(\frac{2}{1.2.3}\) + \(\frac{2}{2.3.4}\) + \(\frac{2}{3.4.5}\) + ... + \(\frac{2}{1998.1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{2.3}\) + \(\frac{1}{2.3}\) - \(\frac{1}{3.4}\) + \(\frac{1}{3.4}\) - \(\frac{1}{4.5}\) + ... + \(\frac{1}{1998.1999}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{1.2}\) - \(\frac{1}{1999.2000}\))
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{3998000}\))
= \(\frac{1}{4}\) - \(\frac{1}{7996000}\) = ?
c) C = \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + \(\frac{1}{6.8}\) + ... + \(\frac{1}{2006.2008}\)
= \(\frac{1}{2}\) (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + \(\frac{1}{2}\)(\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + \(\frac{1}{2}\)(\(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + ... + \(\frac{1}{2006}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\)(\(\frac{1}{2}\) - \(\frac{1}{2008}\))
= \(\frac{1}{2}\) . \(\frac{1003}{2008}\) = \(\frac{1003}{4016}\).
Tính giá trị của biểu thức:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(B=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+....+\frac{1}{990}\)
dễ mà bạn làm từ câu a nếu ra thì các câu khác cũng dễ thôi
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A=1-\frac{1}{2010}\)
\(A=\frac{2009}{2010}\)
K=\(\frac{4}{2.4}\)+\(\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
F=\(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
I=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(K=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(K=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(K=2\times\frac{502}{1005}\)
\(K=\frac{1004}{1005}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+...+\frac{1}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{10}{33}:3\)
\(F=\frac{10}{99}\)
\(I=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(I=1-\frac{1}{2010}\)
\(I=\frac{2009}{2010}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}=\frac{2016}{2017}\)
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{2015.2016}\)
S=2-1/1.2 . 3-2/2.3............2016-2015/2015.2016
=1/1 - 1/2 + 1/2 - 1/3+........+1/2015 - 1/2016
=1/1 - 1/2016
=2015/2016
Tính một cách hợp lí tổng sau :
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}.\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)
\(A=\frac{1}{1}-\frac{1}{2017}\)
\(A=\frac{2016}{2017}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)
\(\Rightarrow A=1-\frac{1}{2017}\)
\(\Rightarrow A=\frac{2016}{2017}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=?\)