Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1.3.5.7...39}{21.22.23...40}=\frac{\left(2.4.6.8...40\right).\left(1.3.5.7...39\right)}{\left(2.4.6.8...40\right).\left(21.22.23...40\right)}=\frac{1.2.3.4...40}{^{2^{20}.1.2.3.4...40}}=\frac{1}{2^{20}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
CM: \(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1}{2^{20}}\)
Biến đổi vế trái:
\(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{22\cdot24\cdot26\cdot\cdot\cdot40}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{2\cdot11\cdot2^3\cdot3\cdot2\cdot13\cdot2^2\cdot7\cdot2\cdot15\cdot2^5\cdot2\cdot17\cdot2^2\cdot9\cdot2\cdot19\cdot2^3\cdot5}\)
\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{\left(3\cdot5\cdot7\cdot\cdot\cdot19\right)2^{20}}\)
\(=\dfrac{1}{2^{20}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhân cả tử và mẫu của phân số\(\frac{1.3.5.....39}{21.22.23.....40}\)với 2.4....40 ta có:
\(\frac{\left(1.3.5.....39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)\(=\frac{1.2.3....39.40}{\left(21.22.23...40\right).\left(1.2.3...20\right)2^{20}}\)
\(=\frac{1.2.3...40}{1.2.3...40.2^{20}}\)\(=\frac{1}{2^{20}}\)
vậy\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\left(đpcm\right)\)
Ta có : \(U=\frac{1.3...39}{21.22...40}\)
=> \(U=\frac{1.3...39.\left(2.4...40\right)}{21.22...40.\left(2.4.6...40\right)}\)
=> \(U=\frac{1.2.3...39.40}{21.22...40.\left(1.2...20\right).2^{20}}\)
=> \(U=\frac{1}{2^{20}}\)
- Ta thấy : \(2^{20}>2^{20}-1\)
=> \(\frac{1}{2^{20}}< \frac{1}{2^{20}-1}\)
hay \(U< V\)
Vậy U < V .
Vì \(^{2^{20}>2^{20}-1}\) chứ