hết cỡ rồi ông tướng

a) Vế trái  \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)

               \(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)

b) Vế trái

 \(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)

ko giải đâu

đùa thôi =)

= \(\frac{1.3.5...19}{22.24....40}\)( triệt tiêu 21 . 23 . 25 ... 39 ) = \(\frac{1.3.5.7...19}{2^{10}.11.12...20}\)=\(\frac{1.3.7.9...19}{2^{15}.6.7.8.9.10}\)=\(\frac{1.3.5}{2^{18}.3.4.5}=\frac{1}{2^{20}}\)

Ta có:\(\frac{1.3.5......39}{21.22.23........4}=\frac{1.3.5....39.2.4.6...40}{21.22.23......40.2.4.6.....40}\)

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

=\(\frac{40!}{40!2^{20}}=\frac{1}{2^{20}}\)

a) \(\frac{1.3.5...39}{21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{\left(2.4.6...40\right).21.22.23...40}=\frac{1.2.3.4.5.6...39.40}{2^{20}.1.2.3...20.21.22.23...40}\)

\(=\frac{1}{2^{20}}\left(đpcm\right)\)

b) \(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{\left(2.4.6...2n\right)\left(n+1\right)\left(n+2\right)...2n}=\frac{1.2.3.4.5.6...\left(2n-1\right).2n}{2^n.1.2.3...n\left(n+1\right)\left(n+2\right)...2n}\)

\(=\frac{1}{2^n}\left(đpcm\right)\)