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

Bạn ơi giản ước thừa số 100 kìa

\(R=1.3.5.7...99\)

\(R=\frac{1.2.3.4.5.6.7.8...99.100}{2.4.6.8...100}\)

\(R=\frac{1.2.3.4.5.6..8...99.100}{\left(2.2.2.2...2\right).\left(1.2.3.4...50\right)}\)

\(R=\frac{51.52.53...100}{2.2.2.2...2}\)

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

Vậy R = S

Lời giải:

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

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

\(A=1.3.5.7...99=\frac{\left(1.3.5.7...99\right)\left(2.4.6...100\right)}{2.4.6...100}=\frac{1.2.3...100}{\left(2.1\right)\left(2.2\right)...\left(2.50\right)}=\frac{\left(1.2.3...50\right)\left(51.52.53....100\right)}{\left(1.2.3...50\right)\left(2.2.2...2\right)}=\frac{51.52.53...100}{2.2...2}=\frac{51}{2}.\frac{52}{2}.\frac{53}{2}...\frac{100}{2}=B\)

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

A > D 

Nha Thị Hồ Lê

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

-->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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