\(\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)

\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-......-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{5}+\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}=\frac{1}{18}\)

x + 1 = 18

x = 18 - 1 = 17         

1/3.4+1/4.5+1/5.6+.....+1/x(x+1)=3/10

1/3-1/4+1/4-1/5+1/5-........-1/x+1/x-1/x+1=3/10

=>1/3-1/x+1=3/10

   1/x+1=3/10-1/3=1/30

=>x+1=30

   x=30-1

   x=29

Ta có :

\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)

=>\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

=>\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

=>\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)

=>\(\frac{1}{x+1}=\frac{1}{30}\)

=>\(x+1=30\)

=>\(x=30-1\)

=>\(x=29\)

Vậy \(x=29\)

Theo đề suy ra

\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)

=> \(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)

\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)

=>x+1=30

=>x=29

\(\text{Ta có: }\frac{1}{5.6}+\frac{1}{6.7}+.....+\frac{1}{x.\left(x+1\right)}=\frac{13}{90}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{\left(x+1\right)}=\frac{13}{90}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 = 18

=> x = 17

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009 }\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\frac{x+1-1}{x+1}=\frac{2008}{2009}\)
\(\frac{x}{x+1}=\frac{2008}{2009}\)
\(2009x=2008\left(x+1\right)\)
\(2009x=2008x+2008\)
\(2009x-2008x=2008\)
\(x=2008\)
Vậy x=2008

Ta có

1/x.(x+1) =2008-1/1.2-1/2.3-....

tự làm nhé!!

=> \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) +...+\(\frac{1}{x\left(x+1\right)}\) = \(\frac{2008}{2009}\)

=> \(\frac{1}{1}\) - \(\frac{1}{2}\) + \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) +...+ \(\frac{1}{x}\) - \(\frac{1}{x+1}\) = \(\frac{2008}{2009}\)

=> \(\frac{1}{1}\) - \(\frac{1}{x+1}\) = \(\frac{2008}{2009}\) => \(\frac{1}{x+1}\) = \(\frac{1}{1}\) - \(\frac{2008}{2009}\) = \(\frac{1}{2009}\) => x+1=2009 => x=2008. Vậy x=2008.