\(\Rightarrow A=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

\(A=4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{95.99}\right)\)

\(A=4.\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=\frac{4}{4}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(A=\frac{32}{99}\)

\(\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+\frac{4}{11}.\frac{4}{15}+...+\frac{4}{95}.\frac{4}{99}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Leftrightarrow A=\frac{32}{99}\)

sai rồi

Đặt \(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+...+\frac{4}{95}\cdot\frac{4}{99}\)

\(A=\frac{16}{21}+\frac{16}{77}+...+\frac{16}{9405}\)

\(A=\frac{16}{3\cdot7}+\frac{16}{7\cdot11}+....+\frac{16}{95\cdot99}\)

\(A=\frac{16}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=4\cdot\left(\frac{1}{3}\cdot\frac{1}{99}\right)=4\cdot\frac{32}{99}=\frac{128}{99}\)

\(A=\frac{4}{3}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{11}+...+\frac{4}{91}\times\frac{4}{95}+\frac{4}{95}\times\frac{4}{99}\)

\(=4\left(\frac{1}{3\times7}+\frac{1}{7.11}+...+\frac{1}{91\times95}+\frac{1}{95\times99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{91}-\frac{1}{95}+\frac{1}{95}-\frac{1}{99}\right)\)

\(=4\left(\frac{1}{3}-\frac{1}{99}\right)=4\times\frac{32}{99}=\frac{128}{99}\)

\(\frac{A}{4}=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\)

\(\frac{A}{4}=\frac{7-3}{3.7}+\frac{11-7}{7.11}+...+\frac{99-95}{95.99}\)

\(\frac{A}{4}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(A=\frac{4.32}{99}\)

\(4.A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{95}-\frac{1}{99}\\ 4.A=\frac{1}{3}-\frac{1}{99}\\ 4.A=\frac{32}{99}\\ A=\frac{32}{99}:4\\ A=\frac{8}{99}\)

\(A=\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\)

\(A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\)

\(A=\frac{1}{3}-\frac{1}{99}\)

\(A=\frac{32}{99}\)

Vậy \(A=\frac{32}{99}\)

\(A=\frac{16}{3.7}+\frac{16}{7.11}+...+\frac{16}{95.99}\)

\(\Leftrightarrow\frac{1}{4}A=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

\(\Rightarrow A=\frac{32}{99}:\frac{1}{4}=\frac{128}{99}\)

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