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\(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+...+\frac{4}{28.30}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{28.10}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+...+\frac{1}{28}-\frac{1}{30}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{30}\right)=2.\left(\frac{15}{60}-\frac{2}{60}\right)=2.\frac{13}{60}=\frac{26}{60}=\frac{13}{30}\)
=5/2(1/4-1/6+1/6-1/8+...+1/208-1/300)
=5/2(1/4-1/300)
=5/2.37/150=37/60
Q=1/4(1.4/2.3+2.5/3.4+3.6/4.5+...+48.51/49.50)
=1/4(2.3−2/2.3+3.4−2/3.4+4.5−2/4.5+...+49.50−2/49.50)
=1/4(1− 2/2.3+ 1− 2/3.4+ 1− 2/4.5+...+1− 2/49.50)
=1/4[48−2(1/2.3+1/3.4+...+1/49.50)]
=1/4[48−2(1/2−1/3+1/3−1/4+...+1/49−150)]
=14[48−2(1/2−1/50)]=294/25
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
\(S=\frac{1}{2}-\frac{1}{10}\)
\(S=\frac{2}{5}\)
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\frac{37}{150}\)
\(=\frac{37}{60}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\frac{37}{150}\)
= \(\frac{37}{60}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
S = \(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
=> 2S = \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\)
=> 2S = \(\frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)
=> S = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{52}=\dfrac{1}{2}-\dfrac{1}{52}=\dfrac{25}{52}\)
$a)\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}$
$=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}$
$=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}$
$=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}$
b) Đặt $B=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{18.20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{18}-\dfrac{1}{20}$
$=>\dfrac{1}{2}B=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}$
$=>B=\dfrac{9}{10}$
c) $\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}$
$=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}$
$=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}$
$=1-\dfrac{1}{9}=\dfrac{8}{9}$
d) Viết lại đề rõ ràng nha bạn.
\(\dfrac{4}{6.8}+\dfrac{4}{8.10}+...+\dfrac{4}{40.42}\)
\(=2\left(\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{40.42}\right)\)
\(=2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{42}\right)\)
\(=2\left(\dfrac{1}{6}-\dfrac{1}{42}\right)\)
\(=2.\dfrac{1}{7}\)
\(=\dfrac{2}{7}\)
Cho \(A=\dfrac{4}{6.8}+\dfrac{4}{8.10}+...+\dfrac{4}{40.42}\)
\(2A=\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{40.42}\)
\(2A=\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{42}\)
\(2A=\dfrac{1}{6}-\dfrac{1}{42}\)
\(2A=\dfrac{1}{7}\)
\(A=\dfrac{1}{7}:2\)
\(A=\dfrac{1}{14}\)