1 và 2/15

Ta có: \(\frac{15}{16}\times\frac{24}{25}\times\ldots\times\frac{9999}{10000}\)

\(=\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)\times\ldots\times\left(1-\frac{1}{10000}\right)\)

\(=\left(1-\frac14\right)\times\left(1-\frac15\right)\times\ldots\times\left(1-\frac{1}{100}\right)\times\left(1+\frac14\right)\times\left(1+\frac15\right)\times\ldots\times\left(1+\frac{1}{100}\right)\)

\(=\frac34\times\frac45\times\ldots\times\frac{99}{100}\times\frac54\times\frac65\times\ldots\times\frac{101}{100}\)

\(=\frac{3}{100}\times\frac{101}{4}=\frac{303}{400}\)

\(\Rightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{99.101}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)

\(\Rightarrow A=\frac{44}{303}\)

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

=> A = 98/203 : 2 = 49/303

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{63}+\frac{100}{99}\\ =\frac{2.2}{1.3}+\frac{4.4}{3.5}+\frac{6.6}{5.7}+\frac{8.8}{7.9}+\frac{10.10}{9.11}\)

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{65}+\frac{100}{99}\)

\(1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+1+\frac{1}{65}+1+\frac{1}{99}\)

\(\left(1+1+1+1+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{65}+\frac{1}{99}\right)\)

\(\frac{60}{11}\)

A. 36,49,64

A đúng ko

A=1/3.5+1/5.7+1/7.9+...+1/99.101

2A= 2/3.5+2/5.7+2/7.9+...+2/99.101

2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101

2A=1/3-1/101=98/303

A=(98/303)/2=49/303

A=97

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