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 .

\(\frac{1.3.5.7...39}{21.22.23...40}=\frac{\left(2.4.6.8...40\right).\left(1.3.5.7...39\right)}{\left(2.4.6.8...40\right).\left(21.22.23...40\right)}=\frac{1.2.3.4...40}{^{2^{20}.1.2.3.4...40}}=\frac{1}{2^{20}}\)

\(\frac{1.3.5.7....39}{21.22.23....40}=\frac{\left(2.4.6....40\right).\left(1.3.5.7....39\right)}{\left(2.4.6....40\right).\left(21.22.23...40\right)}=\frac{1.2.3.4....40}{2^{20}.1.2.3.4....40}=\frac{1}{2^{20}}\)

ko giải đâu

đùa thôi =)

CM: \(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1}{2^{20}}\)

Biến đổi vế trái:

\(\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot39}{21\cdot22\cdot23\cdot\cdot\cdot40}=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{22\cdot24\cdot26\cdot\cdot\cdot40}\)

\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{2\cdot11\cdot2^3\cdot3\cdot2\cdot13\cdot2^2\cdot7\cdot2\cdot15\cdot2^5\cdot2\cdot17\cdot2^2\cdot9\cdot2\cdot19\cdot2^3\cdot5}\)

\(=\dfrac{1\cdot3\cdot5\cdot7\cdot\cdot\cdot19}{\left(3\cdot5\cdot7\cdot\cdot\cdot19\right)2^{20}}\)

\(=\dfrac{1}{2^{20}}\)

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\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\left(đpcm\right)\)