CMR \(1.3.5.....99=\frac{11}{2}.\frac{12}{2}.\frac{13}{2}.....\frac{100}{2}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{11}{2}.\frac{12}{2}.\frac{13}{2}....\frac{20}{2}=\frac{11.12.13....20}{2^{10}}=\frac{11.13.15....19.\left(12.14.16.18.20\right)}{2^{10}}=\)
\(=\frac{11.13.15....19.\left(\left(3.2^2\right).\left(7.2\right).\left(2^4\right).\left(9.2\right).\left(5.2^2\right)\right)}{2^{10}}=\frac{11.13.15....19.\left(3.5.7.9\right).2^{10}}{2^{10}}=\)
\(=1.3.5.7.9.11.13.15....19\left(Đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(A=1.3.5.....99=\frac{1.2.3.....100}{2.4.6.....100}=\frac{1.2.3.....100}{2.\left(2.2\right).\left(2.3\right).....\left(2.50\right)}\)
=> \(A=\frac{51.51.53.....100}{2.2.2....2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.....\frac{100}{2}\)
Vậy ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.3.5....99=\frac{1.2.3.4....99.100}{2.4.6...100}=\frac{\left(1.2.3....50\right).\left(51.52.53...100\right)}{2^{50}.\left(1.2.3...50\right)}\)
\(=\frac{51.52.53....100}{2^{50}}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}......\frac{100}{2}\)
Ta có :
\(1.3.5.....99=\frac{1.2.3.4.....99.100}{2.4.6......100}\)
\(=\frac{1.2.3......99.100}{1.2.2.2.2.3......2.50}\)
\(=\frac{1.2.3.4......99.100}{2^{50}.1.2.3......50}\)
\(=\frac{51.52.....100}{2^{50}}\)
\(=\frac{51}{2}.\frac{52}{2}...........\frac{100}{2}\) (ĐPCM)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Tính:
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{3}{1.2}-\frac{5}{2.3}+\frac{7}{3.4}-\frac{9}{4.5}+\frac{11}{5.6}-\frac{13}{6.7}+\frac{15}{7.8}-\frac{17}{8.9}+\frac{19}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{11}{2}.\frac{12}{2}.\frac{13}{2}...\frac{20}{2}\)
\(=\frac{\left(11.13.15.17.19\right).12.14.16.18.20}{2^{10}}\)
\(\frac{\left(11.13.15.17.19\right).\left(3.2^2\right).\left(7.2^1\right).2^4.\left(9.2^1\right).\left(5.2^2\right)}{2^{10}}\)
\(=\frac{\left(1.3.5.7.9.11.13.15.17.19\right).2^{10}}{2^{10}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)