mai tau giải cho dừ viết lâu lắm. Đúng là phải thưởng.

1.3.5. ... .99=51/2.52/2. ... .100/2
nhân cả hai vế với 1.2...50.2^50, ta được
*vế 1
1.3.5. ... .99.1.2...50.2^50=1.3.5...99.2.2.2..2..1.2...50
=1.3.5...99.1.2.2.2.2.3.2.4.....2.50
1.3.....99.2.4..10=1.2.3.4.5...100 (1)
*vế 2
51/2.52/2. ... .100/2^50.1.2.3...50=51/2.52/2. ... .100/2.2.2...1.2.3...50
=(51/2).2.(52/2).2 ... .(100/2).2.....1.2.3...50
rút gọn ta sẽ đươc51.52.53...100.1.2.3...50(2)
từ (1) và (2)=>1.3.5. ... .99=51/2.52/2. ... .100/2

\(1.3.5....99=\frac{1.2.3.4....99.100}{2.4.6...100}=\frac{\left(1.2.3....50\right).\left(51.52.53...100\right)}{2^{50}.\left(1.2.3...50\right)}\)

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

Ta có :

\(1.3.5.....99=\frac{1.2.3.4.....99.100}{2.4.6......100}\)

\(=\frac{1.2.3......99.100}{1.2.2.2.2.3......2.50}\)

\(=\frac{1.2.3.4......99.100}{2^{50}.1.2.3......50}\)

\(=\frac{51.52.....100}{2^{50}}\)

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

sfdsa

VÌ 1/1.1/3.......1/99=2/51.2/52.........2/100

VÀ   2/51.2/52.....2/100=1/1.1/3.......1/99

SUY RA BẰNG NHAU