Lời giải:

\(M=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2).(n+1)(n+2)....2n}=\frac{(2n-1)!}{2.1.2.2.2.3...2(n-1).(n+1).(n+2)...2n}\)

\(=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).(n+1).(n+2)....2n}=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).n(n+1)..(2n-1).2}\)

\(=\frac{(2n-1)!}{2^{n-1}.(2n-1)!.2}=\frac{1}{2^{n-1}.2}<\frac{1}{2^{n-1}}\)

Ta có đpcm.

Ta có: \(\frac{1.3.5.7.....\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}\)

\(=\frac{1.2.3.4..5.6...\left(2n-1\right).2n}{\left(2.4.6....2n\right)\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)

\(=\frac{1.2.3.4.5.6...\left(2n-1\right)}{2^n.1.2.3....n\left(n+1\right)\left(n+2\right)\left(n+3\right)....2n}\)

\(=\frac{1}{2^n}\left(đpcm\right)\)

Ta có:

\(1.3.5.7.9...\left(2n-1\right)=\frac{\left[1.3.5.7.9....\left(2n-1\right)\right].\left[2.4.6.8...2n\right]}{2.4.6.8....2n}=\frac{1.2.3.4.5.6....2n}{\left(2.1\right).\left(2.2\right).\left(2.3\right)\left(2.4\right)....\left(2.n\right)}\)

=> \(1.3.5.7.9...\left(2n-1\right)=\frac{1.2.3.4.5.6....2n}{\left(2.2.2.....2\right).\left(1.2.3.4.....n\right)}=\frac{\left(1.2.3.4.....n\right)\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}{2^n.\left(1.2.3.4....n\right)}\)

=> \(1.3.5.7.9...\left(2n-1\right)=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n}\)

=> \(\frac{1.3.5.7.9...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}=\frac{\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n}{2^n\left[\left(n+1\right)\left(n+2\right)\left(n+3\right).....2n\right]}=\frac{1}{2^n}\)(đpcm)

a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :

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

\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)

b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :

\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)

\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)

a, 59x + 46y = 2004

Vì 2004 là số chẵn, 46y là số chẵn => 59x là số chẵn

=> x là số chẵn, mà x là số nguyên tố

=> x = 2

=> 2.59 + 46y = 2004

=> 46y = 2004 ‐ 118

=> 46y = 1886

=> y = 1886:46 => y = 41

Vậy x = 2; y = 41

đã làm đề 23 rùi hả!!!!!