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\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{2}-\frac{1}{12}\)
\(=\frac{5}{12}\)
bn sẽ tinh theo kieeuranhaan 2 nha xin lỗi mik làm bi này rùi nhưng mik quên mik có sacks xem lại
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)
Chỉ cần viết ra là: \(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}=1-\frac{1}{11}=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}\)\(+\frac{1}{110}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...\) \(+\frac{1}{9\cdot10}\)\(+\frac{1}{10\cdot11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\)\(\frac{1}{5}\)\(+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\)\(+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(A=\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}\)
\(A=\frac{1}{4\times5}+\frac{1}{5\times6}+...+\frac{1}{11\times12}\)
\(A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(A=\frac{1}{4}-\frac{1}{12}\)
\(A=\frac{3}{12}-\frac{1}{12}=\frac{2}{12}=\frac{1}{6}\)
9/1-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2=0/4
Giải :
ta có
\(\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
=\(\frac{9}{10}-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{9\times10}\right)\)
=\(\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{9}{10}-\left[1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\right]\)
=\(\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
=\(\frac{9}{10}-1+\frac{1}{10}=0\) (Mong online math ks cho mình nhé)
Tính tổng
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.........+\frac{1}{110}+\frac{1}{132}\)
=1/1*2+1/2*3+1/3*4+...+1*10*11+1/11*12=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11+1/11-1/12
=1-1/12=11/12.
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{10\times11}+\frac{1}{11\times12}\)
\(=1-\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{11}+\frac{1}{12}\)
\(=1-\frac{1}{12}\)
\(=\frac{11}{12}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-......-\left(\frac{1}{6}-\frac{1}{6}\right)-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
1/2+1/6+1/12+1/20+1/30+1/42
=1/1x2+1/2x3+1/3x4+1/4x5+1/5x6+1/6x7
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{56}=\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}...\frac{1}{7x8}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)\(-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
b,
\(A=1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)
\(=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1+\frac{1}{2}-\frac{1}{11}=\frac{31}{22}\)
\(A=1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=1+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(A=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=1+\frac{1}{2}-\frac{1}{11}\)
\(A=\frac{31}{22}\)
Vậy \(A=\frac{31}{22}\)