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\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
Ta có: 1/1.2 + 1/2.3 +1/3.4 +......+1/998.999 + 1/999. 1000
= 1/2 + 1/6 + 1/12 + .... + 1/997002 + 1/999000
lại có : 1/2 = 1-1/2
1/6 = 1/2 -1/3
1/12 = 1/3 - 1/4
...
1/997002 = 1/998 - 1/999
1/999000 = 1/999 - 1000
=>1/1.2 + 1/2.3 +1/3.4 +......+1/998.999 + 1/999. 1000
= 1-1/2 + 1/2 - 1/3 + 1/3 -1/4 +....+ 1/998 - 1/999 + 1/999 - 1/1000
= 1-1/1000
= 999/1000
a) \(C=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=7\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=7\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=7.\frac{13}{28}=\frac{7.13}{28}=\frac{13}{4}\)
b) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{999\cdot1000}+1\)
\(=\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+...+\frac{1000-999}{999\cdot1000}+1\)
\(=\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+...+\frac{1000}{999\cdot1000}-\frac{999}{999\cdot1000}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{999}-\frac{1}{1000}+1\)
\(=1-\frac{1}{1000}+1\)
\(=\frac{999}{1000}+1\)
\(=\frac{1999}{1000}\)
giai câu a
a) ta có (\(\frac{2}{11.13}\)+\(\frac{2}{13.15}\)+.....+\(\frac{2}{19.21}\))*462 - x =19
(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{19}\)-\(\frac{1}{21}\)) * 462 -x=19
(\(\frac{1}{11}\)-\(\frac{1}{21}\))*462-x=19
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
\(\frac{23}{12}\)
\(\frac{314}{105}\)
\(\frac{59}{60}\)
\(\frac{199}{90}\)
\(\frac{1}{18}\)
\(\frac{13}{36}\)
\(\frac{4}{221}\)
\(\frac{4}{85}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...+\frac{1}{2014}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
\(B=\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{97.99}=\frac{2}{11}-\frac{2}{13}+...+\frac{2}{97}-\frac{2}{99}=\frac{2}{11}-\frac{2}{99}=\frac{16}{99}\)
C=(1x2)+(2x3)+...+(99x100)
3xC=1x2x3+2x3x3+...+3x99x100
3xC=1x2x3-0x1x2+2x3x4-1x2x3+...+99x100x101-98x99x100
3xC=99x100x101
3xC=999900
C=333300