\(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)
\(S=\dfrac{67}{110}\)
bn có nhầm \(\dfrac{1}{2}\) với \(\dfrac{1}{20}koz\)
\(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)
\(S=\dfrac{7}{10}-\dfrac{1}{11}\)
\(S=\dfrac{67}{110}\)
Mk trình bày lại, cái ở trên bị lỗi nha 
Giải xong mới thấy, cậu viết sai đề rùi 
, \(\dfrac{1}{20}\) sao? phải là \(\dfrac{1}{20}\) mới đúng chứ nhỉ?
Nếu \(S=\dfrac{1}{2}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\) thì:
\(S=\dfrac{1}{2}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(S=\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{11}\)
\(S=\dfrac{55}{110}+\dfrac{22}{110}-\dfrac{10}{110}\)
\(S=\dfrac{67}{110}\)
Nếu \(S=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\) thì:
\(S=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}\)
\(S=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(S=\dfrac{1}{4}-\dfrac{1}{11}\)
\(S=\dfrac{11}{44}-\dfrac{4}{44}\)
\(S=\dfrac{7}{44}\)
(Mình làm thế này vì cảm thấy đề bạn bị sai nhé).
Đề bị nhầm hay gì ấy bạn, theo đúng đề phải là:
\(S=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
Giải:
\(S=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}\)
\(\Rightarrow S=\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}\)
\(S=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\)
\(S=\dfrac{1}{4}-\dfrac{1}{11}\)
\(S=\dfrac{7}{44}\)
Trần Lê Việt Hoàng
hình như bn thiếu số "0"ở phân số \(\dfrac{1}{2}\)đó!
XEM LẠI BÀI GIÙM NHÉ BN !!
S = \(\dfrac{1}{2}\)+ \(\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)+ \(\dfrac{1}{56}\)+ \(\dfrac{1}{72}\)+ \(\dfrac{1}{90}\)+ \(\dfrac{1}{110}\)
S= \(\dfrac{1}{2}\)+ \(\dfrac{1}{5.6}\)+ \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\)+ \(\dfrac{1}{8.9}\)+ \(\dfrac{1}{9.10}\)+ \(\dfrac{1}{10.11}\)
S= \(\dfrac{1}{2}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\)- \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\)- \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)- \(\dfrac{1}{9}\)+ \(\dfrac{1}{9}\)- \(\dfrac{1}{10}\)+ \(\dfrac{1}{10}\)-\(\dfrac{1}{11}\)
S= \(\dfrac{1}{2}\)+ \(\dfrac{1}{5}\)- \(\dfrac{1}{11}\)
S= \(\dfrac{7}{10}\)- \(\dfrac{1}{11}\)
S= \(\dfrac{67}{110}\)
