= \(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+ ... + \(\frac{1}{10.11}\)
= \(\frac{1}{1}\)-\(\frac{1}{2}\)+ \(\frac{1}{2}\)-\(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+... + \(\frac{1}{10}\)-\(\frac{1}{11}\)
= \(\frac{1}{1}\)-\(\frac{1}{11}\)
= \(\frac{10}{11}\)
ai tốt bụng thì tk mk nha, mk đg âm điểm đây
tôi nghĩ sai đề\(\frac{1}{8}\)thành \(\frac{1}{6}\)chứ
ừm chắc mình ghi nhầm đổi \(\frac{1}{8}\)thành \(\frac{1}{6}\)nha sorri các bn
\(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{1}-\frac{1}{11}=\frac{10}{11}\)
\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+...+\(\frac{1}{110}\)
=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{10.11}\)
=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{10}\)-\(\frac{1}{11}\)
=1+(\(\frac{-1}{2}\)+\(\frac{1}{2}\))+(\(\frac{-1}{3}\)+\(\frac{1}{3}\))+(\(\frac{-1}{4}\)+\(\frac{1}{4}\))+...+(\(\frac{-1}{10}\)+\(\frac{1}{10}\))-\(\frac{1}{11}\)
=1+0+0+0+...+0-\(\frac{1}{11}\)
=1-\(\frac{1}{11}\)
=\(\frac{11}{11}\)-\(\frac{1}{11}\)
=\(\frac{10}{11}\)