bang nhau

Giai:

A=1.3.5.7...97.99=\(\frac{\left(1.3.5...97.99\right).\left(2.4.6...100\right)}{2.4.6...100}\)

=\(\frac{1.2.3.4...99.100}{\left(1.2\right).\left(2.2\right)...\left(2.50\right)}\)

=\(\frac{\left(1.2.3...50\right).\left(51.52...99.100\right)}{\left(1.2.3...49.50\right).2^{50}}\)

=\(\frac{51.52...99.100}{2.2...2.2}\)

=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)

mà B=\(\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}\)

Nên A=B

Vậy A=B

\(1.3.5.7...97.99=\frac{100!}{2.4.6.8...100}\)

\(=\frac{1.2.3.4...100}{1.2.2.2.3.2...50.2}\)

\(=\frac{51.52.53...100}{2}\)

Vậy \(A=B\)

\(A=1.3.5.....97.99\)

\(=\frac{1.2.3.4......98.99.100}{2.4.6.8......96.98.100}\)

\(=\frac{1.2.3.4....98.99.100}{2.1.2.2.2.3........49.2.50.2}\)

\(=\frac{\left(1.2.3.4......50\right)51.52....98.99.100}{2^{50}\left(1.2.3.......50\right)}\)

\(=\frac{51.52.53.....99.100}{2^{50}}\)

\(=\frac{51}{2}\cdot\frac{52}{2}\cdot\frac{53}{2}\cdot.......\cdot\frac{99}{2}\cdot\frac{100}{2}=B\)

Vậy \(A=B\)

Ta có:

\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}=\dfrac{51.52.53...100}{2^{50}}\)

\(=\dfrac{\left(51.52.53..100\right)\left(1.2.3.4...50\right)}{2^{50}\left(1.2.3.4...50\right)}\)

\(=\dfrac{1.2.3.4.5.6...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)

\(=\dfrac{1.2.3.4.5.6...100}{2.4.6.8.10...100}=\dfrac{\left(1.3...99\right)\left(2.4...100\right)}{2.4.6...100}\)

\(=1.3.5.7.99=A\)

Vậy \(A=B\) (Đpcm)

\(R=1.3.5.7...99\)

\(R=\frac{1.2.3.4.5.6.7.8...99.100}{2.4.6.8...100}\)

\(R=\frac{1.2.3.4.5.6..8...99.100}{\left(2.2.2.2...2\right).\left(1.2.3.4...50\right)}\)

\(R=\frac{51.52.53...100}{2.2.2.2...2}\)

\(R=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=S\)

Vậy R = S

Lời giải:

\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)

\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)

Đặt \(A=1.3.5.7...99\)

\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\)

Ta có:

\(A=1.3.5.7...99\)

\(\Rightarrow A=\dfrac{\left(1.3.5.7...99\right)\left(2.4.6.8...100\right)}{2.4.6.8...100}\)

\(\Rightarrow A=\dfrac{1.2.3.4...100}{2.4.6.8...100}\)

\(\Rightarrow A=\dfrac{1.2.3.4...100}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.50\right)}\)

\(\Rightarrow A=\dfrac{\left(1.2.3.4...50\right)\left(51.52.53...100\right)}{\left(1.2.3.4...50\right)\left(2.2.2.2...2\right)}\)

\(\Rightarrow A=\dfrac{51.52.53.54...100}{2.2.2.2...2}\)

\(\Rightarrow A=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}....\dfrac{100}{2}\)

\(\Rightarrow A=B\)

Vậy \(1.3.5.7...99=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}...\dfrac{100}{2}\) (Đpcm)

VT: 1.3.5.7....99=\(\dfrac{(1.3.5.7.....99).\left(2.4.6....100\right)}{2.4.6....100}\)

\(=\dfrac{\left(1.3.5.7.....99\right)\left(2.4.6.....100\right)}{1.2.2.2.2.3.....2.50}\)\(=\dfrac{\left(1.2.3.4.....50\right)\left(51.52.53....100\right)}{\left(1.2.3.4......50\right)\left(2.2.2.2.2....2\right)}\)

\(=\dfrac{51.52.53......100}{2.2.2.2.....2}=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}......\dfrac{100}{2}=VP\left(đpcm\right)\)

\(A=1.3.5.7...99=\frac{\left(1.3.5.7...99\right)\left(2.4.6...100\right)}{2.4.6...100}=\frac{1.2.3...100}{\left(2.1\right)\left(2.2\right)...\left(2.50\right)}=\frac{\left(1.2.3...50\right)\left(51.52.53....100\right)}{\left(1.2.3...50\right)\left(2.2.2...2\right)}=\frac{51.52.53...100}{2.2...2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=B\)

\(1.3.....99=\frac{1.3....99.2.4.6....100}{2.4.6....100}\)

\(=\frac{1.2.3.4.5......99.100}{2^{50}.\left(1.2.3....50\right)}\)

\(=\frac{51.52.53...100}{2.2.2...2}\)

\(=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\)

\(\Rightarrow1.3...99=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}\left(đpcm\right)\)

Ta có  :\(\frac{51}{2}\) . \(\frac{52}{2}\) .... \(\frac{100}{2}\)

       =\(\frac{51.52....100}{2.2....2}\)

     =\(\frac{51.52....100}{2.2....2}\) . \(\frac{2.4.6....100}{2.4.6....100}\) 

    =\(\frac{51.52....100.2.4.6...100}{2.4.6...100.2.2...2}\)

    =\(\frac{1.2.3.4...100}{2.4.6...100}\)

   =\(\frac{\left[1.3.5....99\right].\left[2.4.6...100\right]}{2.4.6...100}\)

  =1.3.5...99[đpcm]