Question

[trần sơn]

\(\frac{1\cdot3\cdot5\cdot...\cdot39}{21\cdot22\cdot23\cdot...\cdot40}\)=\(\frac{1}{2^{20}}\) 

Answer 1

[KAl(SO4)2·12H2O]

\(\frac{1.3.5....39}{21.22.23.....40}=\frac{\left(1.3.5.....39\right).\left(2.4.6.....40\right)}{\left(21.22.....40\right).\left(2.4.6.....40\right)}\)

\(=\frac{1.2.3.4.....40}{21.22.....40.\left(1.2.3.....40\right)2^{20}}\)

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

Answer 2

[trần sơn]

ở dấu = tú 2 phía trên 40 lấy đâu ra vậy