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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\)
\(=\frac{1}{4}-\frac{1}{50}\)
\(=\frac{23}{100}\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{100}{2}=50\)
\(\frac{-17}{2.4}-\frac{17}{4.6}-\frac{17}{6.8}-...-\frac{17}{100.102}\)
\(=-\frac{17}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{100.102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}\left(\frac{1}{2}-\frac{1}{102}\right)\)
\(=-\frac{17}{2}.\frac{25}{51}=-\frac{25}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}\)
\(=\frac{6}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
2)\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2\times\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
tự làm tiếp nhé
1.= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
= \(1-\frac{1}{101}\) = \(\frac{100}{101}\)
2.= \(2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
= \(2\cdot\left(\frac{1}{2}-\frac{1}{2010}\right)\) = \(2\cdot\frac{502}{1005}\) = \(\frac{1004}{1005}\)
\(=\left(1+3+5+...+99+101\right)-\left(2+4+6+...98+100\right)\)
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
2/2*4 + 2/4*6 + 3/6*8 + ... + 2/38*50
= 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + .... + 1/38 - 1/50
= 1/2 - 1/50
= 24/50
= 12/25