Question

Tính giá trị của biểu thức: \(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)

Answer 1

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

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

\(B=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

Vậy giá trị của biểu thức \(B=\frac{32}{99}\)

Answer 2

Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)

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

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

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

Answer 3

Ta Có

(1/3-1/7+1/7-1/11+1/11-1/15+...+1/95-1/99)

(1/3-1/99)

32/99

Answer 4

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

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

Answer 5

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

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

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

 B = \(\frac{1}{3}-\frac{1}{99}\)

 B = \(\frac{33}{99}-\frac{1}{99}\)

 B = \(\frac{32}{99}\)