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:\(B=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}.\frac{54}{2}.....\frac{99}{2}.\frac{100}{2}\)
\(B=\frac{51.52.53.54.....99.100}{2^{50}}\)
\(B=\frac{\left(51.52.53.54.....99.100\right).1.2.3.4.5.6.....49.50}{2^{50}.1.2.3.4.5.6.....49.50}\)
\(B=\frac{1.2.3.4.5.6.....49.50.51.52.53.54.....99.100}{\left(1.2\right).\left(2.2\right).\left(3.2\right).\left(4.2\right).....\left(49.2\right).\left(50.2\right)}\)
\(B=\frac{\left(1.3.5.....99\right).\left(2.4.6.....98.100\right)}{2.4.6.8.....98.100}\)
\(B=1.3.5.....99=A\)
Vậy B=A
\(A=1.3.5....97.99\)
\(=\frac{1.2.3....98.99.100}{2.4.6....96.98.100}\)
\(=\frac{1.2.3...98.99.100}{1.2.2.2.2.3...49.2.50.2}\)
\(=\frac{51}{2}\cdot\frac{52}{2}\cdot...\cdot\frac{99}{2}\cdot\frac{100}{2}=B\)
\(\Rightarrow A=B\)
