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\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}\)
\(=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
= \(\frac{5}{2}-\frac{5}{4}+\frac{5}{4}-\frac{5}{6}+...+\frac{5}{98}-\frac{5}{100}\)
= \(\frac{5}{2}-\frac{5}{100}\)
= \(\frac{49}{50}\)
\(Q=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{98.100}\)
\(=5\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.2.\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{5}{2}.\frac{49}{100}=\frac{49}{40}\)
\(\Rightarrow Q=\frac{49}{40}\)
\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{48.50}\)
\(A=\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48.50}\right)\)
\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(A=\frac{6}{5}\)
=\(\frac{1}{5}.\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{48}-\frac{1}{50}\right)\right)\)
=\(\frac{1}{5}.\left(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\right)\)
=\(\frac{1}{5}.\left(\frac{1}{2}.\frac{12}{25}\right)\)
=\(\frac{1}{5}.\frac{6}{25}=\frac{6}{125}\)
Vậy \(A=\frac{6}{125}\)
\(=\left(1+3+5+...+99+101\right)-\left(2+4+6+...98+100\right)\)
Thấy từ 1 đến 100 có (101-1)/2+1=51
=> 1+3+5+....+99+100=(1+101)x50/2=2601
Từ 2 đến 100 có (102-2)/2+1=50
=> 2+4+...+98+100=(2+100)X50/2=2550
=> D=2601-2550=51
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+....+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)=\frac{5}{2}.\frac{37}{150}=\frac{37}{60}\)
\(\frac{5}{4\cdot6}+\frac{5}{6\cdot8}+\frac{5}{8\cdot10}+...+\frac{5}{298\cdot300}\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\left(\frac{1}{4}-\frac{1}{300}\right)\)
\(=\frac{5}{2}\cdot\frac{37}{150}\)
\(=\frac{37}{60}\)
\(\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)
= \(\frac{5}{2}.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+...+\frac{2}{298.300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)
= \(\frac{5}{2}.\frac{37}{150}\)
= \(\frac{37}{60}\)
\(=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\)
\(=\frac{1}{4}-\frac{1}{50}\)
\(=\frac{23}{100}\)
Nếu bn hỏi cái 1/4,1/6 các kiểu ở đâu ra thì nó là phân tích của 2/4.6 đấy
Chúc bn hok tốt
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
= \(\dfrac{2}{2}.\left(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+....+\dfrac{5}{48.50}\right)\)
\(\)\(=\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+....+\dfrac{2}{48.50}\right)\)
\(=\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\dfrac{12}{25}\)
=\(\dfrac{6}{5}\)=\(1\dfrac{1}{5}\)
Nếu bạn không biết cách giải bài này có thể bảo mình viết cách giải giúp!!!
Chúc bạn làm tốt!!!
\(\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{48.50}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{48.50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{48}\right)\)
=\(\dfrac{5}{2}.\dfrac{23}{48}\) = \(\dfrac{115}{96}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)