Sửa đề: \(\dfrac{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\)

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

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

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

=1

Đặt C=1.3.5.7...99
Đặt D=51/2.52/2.53/2 ....100/2
Ta có:C=1.3.5.7...99
=>2.4.6...100.C=1.2.3...100
=>C = (1.2.3....100) / (2.4.6...100)= (1.2.3...50).(51.52...100) / [(2.1)(2.2).(2.3)...(2.50)]
C=(1.2.3...50).(51.52...100) /[2^50.(1.2.3...50)] =(51.52...100)/2^50 =51/2.52/2.53/2...100/2 =D
Vậy C=D

Ta có :

\(1.3.5.....99=\frac{\left(1.3.5.....99\right)\left(2.4.6.....98\right)}{2.4.6.....98}=\frac{1.2.3.....99.100}{2^{50}\left(1.2.3.....50\right)}=\frac{51.52.53.....100}{2.2.2.....2}\)

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

Vậy......................

~ Hok tốt ~

Ta có \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

\(\Rightarrow\text{Đ}PCM\)

Ta có:(1+1/3+1/5+...+1/99)-(1/2+1/4+1/6+...+1/100)

       =(1+1/2+1/3+1/4+...+1/99+1/100)-2.(1/2+1/4+...+1/100)

       =(1+1/2+1/3+1/4+...+1/99+1/100)-(1+1/2+1/3+...+1/50)

       =1/51+1/52+...+1/99+1/100(đpcm)

Thanks Giang

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

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

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