\(\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99.101}\right)\)

\(=\left(\frac{3}{3}+\frac{1}{3}\right)\times\left(\frac{8}{8}+\frac{1}{8}\right)\times\left(\frac{15}{15}+\frac{1}{15}\right)\times...\times\left(\frac{9999}{9999}+\frac{1}{9999}\right)\)

\(=\frac{4}{3}\times\frac{9}{8}\times\frac{16}{15}\times...\times\frac{10000}{9999}\)

\(=\frac{4\times9\times16\times...\times10000}{3\times8\times15\times...\times9999}\)

\(=\frac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)

\(=\frac{2\times100}{101}=\frac{200}{101}\)

mk cx co dap an vay

$\dfrac{1}{2+3}$

\(B=\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99\times101}\right)\)

\(B=\frac{1\times3+1}{1\times3}\times\frac{2\times4+1}{2\times4}\times\frac{3\times5+1}{3\times5}\times...\times\frac{99\times101+1}{99\times101}\)

\(B=\frac{2\times2}{1\times3}\times\frac{3\times3}{2\times4}\times\frac{4\times4}{3\times5}\times...\times\frac{100\times100}{99\times101}\)

\(B=\frac{\left(2\times3\times4\times...\times100\right)\times\left(2\times3\times4\times...\times100\right)}{\left(1\times2\times3\times...\times99\right)\times\left(3\times4\times5\times...\times101\right)}\)

\(B=\frac{100\times2}{101}=\frac{200}{101}\)

\(A=\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)\cdot...\cdot\left(1+\frac{1}{99\cdot101}\right)\)

\(A=\frac{4}{1\cdot3}\cdot\frac{9}{2\cdot4}\cdot\frac{16}{3\cdot5}\cdot...\cdot\frac{10000}{99\cdot101}\)

\(A=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)\cdot...\cdot\left(100\cdot100\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(99\cdot101\right)}\)

\(A=\frac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\left(3\cdot4\cdot5\cdot...\cdot101\right)}\)

\(A=\frac{100\cdot2}{1\cdot101}\)

\(A=\frac{200}{101}\)

câu này khó thế

cong nhan

Ps cuối hình như có vấn đề..........