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

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

\(=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4...100}\)

\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{10000}\right)\)

\(=\frac{3}{4}.\frac{8}{9}....\frac{9999}{10000}=\frac{101}{200}\)

A = \(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}=\frac{1\cdot3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99\cdot101}{100\cdot100}=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)

B = ( 1- 1/4 )( 1-1/9) ...( 1-1/10000 ) = 3/4 . 8/9 .....9999/100000 ( tương tự A )

a=5051/100 co ma

xin loi ban lam dung roi minh nham

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{2003}-1\right)\)

=\(\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}.....\frac{-2002}{2003}\)

=\(\frac{1}{2003}\)

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

=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

=\(\frac{\left(1.2.3.....99\right)\left(3.4.5.....101\right)}{\left(2.3.4.....100\right)\left(2.3.4.....100\right)}\)

=\(\frac{101}{100.2}\)

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

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

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{99.101}{100.100}\)

\(=\frac{1.2.3...99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)

\(=\frac{1}{100}.\frac{101}{2}\)

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

3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50

Mày hay nhờ mai tao méc thầy

tự làm đi

\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{624}{625}\)

\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{24.26}{25.25}\)

\(=\frac{1.2.3....24}{2.3.4....25}\cdot\frac{3.4.5....26}{2.3.4....25}\)

\(=\frac{1}{25}\cdot\frac{26}{2}=\frac{26}{50}=\frac{13}{25}\)

\(\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{8}\right)\cdot\left(1+\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1+\frac{1}{9999}\right)\)

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

\(=\frac{2.2}{1.3}\cdot\frac{3.3}{2.4}\cdot\frac{4.4}{3.5}\cdot\cdot\cdot\cdot\frac{100.100}{99.101}\)

\(=\frac{2.3.4...100}{1.2.3...99}\cdot\frac{2.3.4...100}{3.4.5...101}\)

\(=\frac{100}{1}\cdot\frac{2}{101}=\frac{200}{101}\)

mk chỉnh lại đề

\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}....\frac{99}{100}\)

\(=\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.\frac{5^2-1}{5^2}....\frac{10^2-1}{10^2}\)

\(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{9.11}{10^2}\)

\(=\frac{2.3.4...9}{3.4.5...10}.\frac{4.5.6...11}{3.4.5...10}\)

\(=\frac{2}{10}.\frac{11}{3}=\frac{11}{15}\)