\(\frac{51.52.53...100}{1.3.5...99}\)

\(=\frac{\left(2.4.6...100\right).\left(51.52.53...100\right)}{\left(2.4.6...100\right).\left(1.3.5...99\right)}\)

\(=\frac{\left(2.4.6...100\right).\left(51.52.53...100\right)}{1.2.3.4.5.6...99.100}\)

\(=\frac{2.4.6...100}{1.2.3...50}\)

\(=\frac{\left(2.2...2\right).\left(1.2.3...50\right)}{1.2.3...50}\)

\(=2.2.2...2\)

\(=2^{50}\)

sao hỏi khó zzzzzzzzậy

bạn viết tất cả các số giống nhau giữa tử và mẫu ra rồi còn bao nhiêu bạn rút gọn

\(\frac{1.3.5....49}{27.28.29...50}=\frac{1.3.5....\left(27.29...49\right)}{\left(27.29...49\right).\left(28.30...50\right)}=\frac{1.3.5....25}{28.30....50}\)=\(\frac{13}{4^32^6.8.16.32}=\frac{13}{2^6.2^6.2^3.2^4.2^5}=\frac{13}{2^{24}}\)

\(2A=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right).2\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)

\(2A=1-\frac{1}{99}\)
\(2A=\frac{98}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)

Ta có \(1.3.5...99=\frac{1.2.3.4.5...100}{2.4.6...100}=\frac{1.2.3.4.5....100}{2^{50}.1.2.3.4...50}=\frac{51.52.53...100}{2^{50}}\left(\text{đpcm}\right)\)

Ta có : \(1.3.5....99=\frac{1.2.3.4.5....100}{2.4.6...100}=\frac{1.2.3.4.5....1000}{2^{50}.1.2.3.4....50}=\frac{51.51.53....100}{2^{50}}\)( đpcm )

lấy 1 chia cho các tổng rồi áp dụng công thức là ra