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Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
= \(1-\frac{1}{12}=\frac{11}{12}\)
ta có:
1/6+1/12+1/20+1/30+.........+1/90+1/110
= 1/2x3+1/3x4+1/4x5+1/5x6+....+1/9x10+1/10x11
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10+1/10-1/11
=1/2-1/11=11/22-2/22=9/22
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{10}-\frac{1}{11}.\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}.\)
Ta có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{110}\)
\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{10\times11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
Đặt A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{110}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{10.11}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
=\(\frac{10}{11}\)
Vậy...
Dấu \(.\)là dấu nhân
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{11}{11}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Chúc bạn học tốt !!!
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + ... + 1/110
B = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + .... + 1/10.110
B = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/10 - 1/11
B = 1 - 1/11
B = 10/11
B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + .... + 1/110
=1/1x2 + 1/2x3 + 1/ 3x4 + 1/4x5 + 1/5 x6 + 1/ 6x 7 +.....+ 1/10 x11
=1-1/2 + 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7 +......+ 1/10 - 1/11
= 1 + (1/2-1/2)+(1/3-1/3)+(1/4-1/4)+(1/5-1/5)+(1/6-1/6)+.....+(1/10-1/10)-1/11
= 1- 1/11=10/11
Vậy B = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + .... + 1/110 = 10/11
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{10\times11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
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