Question

Tính giá trị của biểu thức :

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

Answer 1

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

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

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

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

Answer 2

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

Answer 3

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

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

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

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

Vậy \(A=\frac{32}{99}\)

Answer 4

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

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

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

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

Vậy \(A=\frac{32}{99}\)