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

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

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

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 \(\frac{1.3.5...39}{21.22.23...40}\) ta được:

\(\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}=\frac{1.2.3...39.40}{21.22.23...40.\left[\left(1.2\right).\left(2.2\right)....\left(2.20\right)\right]}\)

\(=\frac{1.2.3...39.40}{21.22.23...40.\left(1.2.3...20\right).2^{30}}=\frac{1.2.3...39.40}{1.2.3...20.21....40.2^{20}}=\frac{1}{2^{20}}\)

Suy ra điều phải chứng minh.

Úi nhầm ở chỗ kia phải là 2²⁰

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

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

Nhân cả tử và mẫu với 2.4.6.....40, ta được:

\(\frac{1.3.5.....39}{21.22.23.....40}=\frac{\left(1.3.5.....39\right)\left(2.4.6.....40\right)}{\left(21.22.23.....40\right)\left(1.2.3.....20\right).2^{20}}=\frac{1}{2^{20}}\left(đpcm\right)\)

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

Nhân cả từ và mẫu với 2 . 4 . 6 ... 40 ta được:

\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right)\left(2.4.6...40\right)}{\left(21.22.23...40\right)\left(2.4.6...40\right)}=\frac{1.2.3.4...39.40}{21.22.23...40.\left(1.2.3...20\right).2^{20}}=\frac{1}{2^{20}}\)(đpcm)

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

sao lai phai nhan voi 2..4..6..40

nhan thêm ma

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

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