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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
\(Q=\frac{1}{4}\left(\frac{1.4}{2.3}+\frac{2.5}{3.4}+\frac{3.6}{4.5}+...+\frac{48.51}{49.50}\right)\)
\(=\frac{1}{4}\left(\frac{2.3-2}{2.3}+\frac{3.4-2}{3.4}+\frac{4.5-2}{4.5}+...+\frac{49.50-2}{49.50}\right)\)
\(=\frac{1}{4}\left(1-\frac{2}{2.3}+1-\frac{2}{3.4}+1-\frac{2}{4.5}+...+1-\frac{2}{49.50}\right)\)
\(=\frac{1}{4}\left[48-2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\right]\)
\(=\frac{1}{4}\left[48-2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\right]\)
\(=\frac{1}{4}\left[48-2\left(\frac{1}{2}-\frac{1}{50}\right)\right]=\frac{294}{25}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)
= \(\frac{5}{2}-\frac{5}{100}\)
= \(\frac{49}{50}\)
\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)
\(\Rightarrow Q=\frac{49}{40}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
\(A=\frac{-1}{2.4}+\frac{-1}{4.6}+\frac{-1}{6.8}+...+\frac{-1}{98.100}\Leftrightarrow.\)\(-2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{98.100}\Leftrightarrow.\)
\(-2A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\Leftrightarrow.\)
\(-2A=\frac{1}{2}-\frac{1}{100}\Leftrightarrow-2A=\frac{49}{100}\Leftrightarrow A=\frac{-49}{200}.\)
ĐÁP SỐ : \(A=\frac{-49}{200}.\)
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\frac{37}{150}\)
\(=\frac{37}{60}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\frac{37}{150}\)
= \(\frac{37}{60}\)
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...........+\frac{1}{98.100}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
cho mình nha!
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{10}\)
\(2S=\frac{1}{2}-\frac{1}{10}\)
\(2S=\frac{2}{5}\)
\(S=\frac{2}{5}:2\)
\(S=\frac{1}{5}\)
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 = \(\frac{2}{5}:2=\frac{2}{5}x\frac{1}{2}=\frac{1}{5}\)
\(\frac{1.4}{4.6}+\frac{2.5}{6.8}+...+\frac{48.51}{98.100}\)
=> \(\frac{1}{4}.\left(\frac{1.4}{2.3}+\frac{2.5}{3.4}+...+\frac{48.52}{49.50}\right)\)
=> \(\frac{1}{4}.\left(\frac{2.3-2}{2.3}+\frac{3.4-2}{3.4}+...+\frac{49.50-2}{49.50}\right)\)
=> \(\frac{1}{4}.\left(1-\frac{2}{2.3}+1-\frac{2}{3.4}+...+1-\frac{2}{49.50}\right)\)
=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{49.50}\right)\right]\)
=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\right]\)
=> \(\frac{1}{4}.\left[48-2.\left(\frac{1}{2}-\frac{1}{50}\right)\right]\)
=> \(\frac{1}{4}.\left[48-2.\frac{12}{25}\right]\)
=> \(\frac{1}{4}.\frac{1176}{25}=\frac{249}{25}\)