-->C=\(\frac{1.2.3.4...99.100}{2.4.6....100}\)-->C=\(\frac{1.2.3...99.100}{\left(2.2....2\right)\left(1.2.3.4.5....50\right)}\)[50 chữ số 2]
-->\(C=\frac{51}{2}.\left(\frac{52}{2}\right)....\left(\frac{100}{2}\right)\)=D vậy C=D
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LI-KE CHO MK NHÉ BN 

so sánh:  C=1.3.5.7...99 với D= 512 .522 .532 ...1002 

xử lí C ta có C=51.52.53.....100/2⁵⁰

ta nhân cả tử và mẫu của C với 1.2.3.........50  thì dc

(1.2.3.4.5.6.........................50).(51.52..............100)

(1.2.3.4...............................50) (2.2...................2) có 50 thừa số 2

tử giữ nguyên xét mẫu ta có (1.2........50).(2.2.......2.2)= (1.2)(2.2)......(50.2)=2.4.6.8......100 vậy triệt tiêu tử cho mẫu thì ta dc c=1.3....97.99

tức C=D

Ta có:

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

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

\(=\frac{\left(51.52.53....100\right)\left(1.2.3.....50\right)}{2^{50}\left(1.2.3.....50\right)}\)

\(=\frac{1.2.3.....100}{\left(2.1\right)\left(2.2\right)\left(2.3\right).......\left(2.50\right)}\)

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

= 1.3.5.....99 = C

Vậy C = D

Đào An Nguyên

A > D 

Nha Thị Hồ Lê

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
VAy C=D

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 \(A=B\)