\(1\cdot3\cdot5\cdot...\cdot99=\dfrac{\left(1\cdot3\cdot5\cdot...\cdot99\right)\cdot\left(2\cdot4\cdot6\cdot...\cdot100\right)}{2\cdot4\cdot6\cdot...\cdot100}\)

\(=\dfrac{1\cdot3\cdot5\cdot...\cdot2\cdot4\cdot6\cdot...\cdot100}{1\cdot2\cdot3\cdot...\cdot50\cdot2\cdot2\cdot...\cdot2}=\dfrac{51}{2}\cdot\dfrac{52}{2}\cdot...\cdot\dfrac{100}{2}\)

- Ta có : `C=51/2 * 52/2 * 53/2* ... * 100/2`

`-> C=(51.52.53...100)/(2^50)`

`-> C=((1.2.3...50).(51.52.53...100))/((1.2.3...50).2^50)`

`-> C=(1.2.3...100)/((1.2).(2.2).(3.2)...(50.2))`

`-> C=(1.2.3...100)/(2.4.6...100)`

`-> C=1.3.5.7...99`

- Từ đó ta có :

`B-C=1.3.5.7...99-1.3.5.7...99=0`

- Vậy `B-C=0`

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

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

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

Lời giải:

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

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

51/2.52/2....100/2=51.52.53...100/(2^50)

=51.52.53...100.(1.2.3.4....50)/(2^50).(1.2.3...50)

=1.2.3.4...100/(1.2)(2.2)(2.3)(2.4)....(2.50)                      (moi thua so 2 nhan voi thua so 1,2,3...)

=1.2.3....100/2.4.6.8...100

=(1.3.5.7....99)(2.4.6.8...100)/2.4.6.8...100

=1.3.5.7.9...99

Ta có: 1.3.5....99
= [(1.3.5...99)(2.4.6.8....100]/(2.4.6....…
= (1.2.3.4....100)/[(1.2).(2.2).(3.2)...(5…
= [(1.2.3...50)(51.52.53...100)]/[(1.2.3..…
= (51.52.53....100)/(2.2.2.2...2)
Từ 2 -> 100(chỉ có các số chẵn) có 50 số (Áp dụng công thức tính số các số 1 dãy = (cuối - đầu)/khoảng cách 1).
=> Trong cụm (2.2.2.2...2) có 50 chữ số 2 (Vì mỗi chẵn số từ 2 -> 100 đều cho 1 số 2) 
Mà từ 51 - > 100 có 50 số
=> (51.52.53....100)/(2.2.2.2...2) = (51/2).(52/2).(53/2)....(100/2) (Vừa đủ) đpcm

Bài này khó hơn mình nghĩ đấy

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

Sao bạn biết được là không có ai trả lời

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Đặt A= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)

= \(\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\right)\)