\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{8x9}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

=\(1-\frac{1}{9}\)

=\(\frac{8}{9}\)

OK XONG NHỚ CHO MIK NHA

\(\frac{1}{1\times2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+.......+\frac{1}{7x8}+\)\(\frac{1}{8x9}\)

=1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)

=1-\(\frac{1}{9}\)

=\(\frac{8}{9}\)

= 9/10

k nha

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_

a, \(=\frac{1}{10}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\left(1-\frac{1}{10}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)\)

\(=\left(\frac{10}{10}-\frac{1}{10}\right)+0+...+0=\frac{9}{10}\)

...

= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10

= 1/1-1/10

= 9/10

minh ko biet xin loi bn nha!

minh ko biet xin loi bn nha!

minh ko biet xin loi bn nha!

minh ko biet xin loi bn nha!

\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(C=1-\frac{1}{2018}\)

\(C=\frac{2017}{2018}\)

\(C=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+.....+\frac{1}{2017x2018}\)

Ta thấy \(\frac{1}{1x2}=\frac{1}{1}-\frac{1}{2}\)

               \(\frac{1}{2x3}=\frac{1}{2}-\frac{1}{3}\)

      .............................................

           \(\frac{1}{2017x2018}=\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2017}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{1}{1}-\frac{1}{2018}\)

\(\Rightarrow C=\frac{2017}{2018}\)

Chúc bạn học tốt nhớ k mình nhá

\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)

\(=1+\left(\frac{2017}{2018}\right)\)

\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)

\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)

\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)

\(=1+\left(1-\frac{1}{2018}\right)\)

\(=1+\frac{2017}{2018}\)

\(=1+\frac{2017}{2018}\)

\(=\frac{4035}{2018}\)

\(S=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

Áp dụng công thức : \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(S=1-\frac{1}{100}=\frac{99}{100}\)

Dap an la 99/100.nho k cho minh.bai giai se gui sau