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\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(2\left(1-\frac{1}{21}\right)\)
=\(\frac{2.20}{21}=\frac{40}{21}\)
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) \(B=\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{97.99}\)
\(=3\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=3.\frac{32}{99}=\frac{3.32}{99}=\frac{32}{33}\)
\(A=\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}\)
\(A=\frac{2\cdot2}{3\cdot5}+\frac{2\cdot2}{5\cdot7}+\frac{2\cdot2}{7\cdot9}+\frac{2\cdot2}{9\cdot11}\)
\(A=2\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=2\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{8}{33}\)
\(A=\frac{16}{33}\)
Ta có:
\(A=\frac{4}{3.5}+\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}\)
\(A=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(A=2\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(A=2\cdot\frac{8}{33}\)
\(A=\frac{16}{33}\)
\(\frac{4.5.6}{14.15.16}\)=\(\frac{1.1.3}{7.3.4}\)=\(\frac{1.1.1}{7.1.4}\)=\(\frac{1}{28}\)
\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
(2/143+2/195+2/255+2/323+2/399).462-y=19
=> 10/231.462-y=19
=> 20-y=19
=> y= 20-19=1
DUYỆT NHA "T I C K"
\(\frac{19}{35}+\frac{15}{19}+\frac{16}{35}+\frac{23}{19}+\frac{4}{15}+\frac{11}{15}\)
\(=\left(\frac{19}{35}+\frac{16}{35}\right)+\left(\frac{15}{19}+\frac{23}{19}\right)+\left(\frac{4}{15}+\frac{11}{15}\right)\)
\(=\frac{35}{35}+\frac{38}{19}+\frac{15}{15}\)
\(=1+2+1\)
\(=4\)
\(\frac{19}{35}+\frac{15}{19}+\frac{16}{35}+\frac{23}{19}+\frac{4}{15}+\frac{11}{15}\)
\(=\left(\frac{19}{35}+\frac{16}{35}\right)+\left(\frac{15}{19}+\frac{23}{19}\right)+\left(\frac{4}{15}+\frac{11}{15}\right)\)
\(=1+2+1\)
\(=4\)
Học tốt
