Question

[_Girl_Nhân_Mã_]

Tính bằng cách thuận tiện nhất:

(1+\(\frac{1}{4}\) ) x (1+\(\frac{1}{8}\) ) x (1+\(\frac{1}{15}\) ) x (1+\(\frac{1}{24}\) ) x ..... x(1+\(\frac{1}{9999}\) )

CÁC BẠN GIÚP MÌNH VỚI.AI NHANH ,LÀM ĐÚNG VÀ ĐẦY ĐỦ THÌ MIK K CHO.Thank you very much!!!

Answer 1

[Duc Loi]

\(\left(1+\frac{1}{4}\right).\left(1+\frac{1}{8}\right).\left(1+\frac{1}{15}\right).\left(1+\frac{1}{24}\right)...\left(1+\frac{1}{9999}\right)\)

\(=\frac{5}{4}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}...\frac{10000}{9999}=\frac{5.9.16.25...10000}{4.8.15.24...9999}=\frac{5.3^2.4^2.5^2...100^2}{4.2.4.3.5.4.6...99.101}\)

\(=\frac{5.3.4.5...100.3.4.5...100}{4.2.3.4...99.4.5.6...101}=\frac{5.100.3}{4.2.101}=\frac{5.25.3}{2.101}=\frac{375}{202}.\)