Question

tính tổng

A=\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{29.30}\)

B=\(\dfrac{4}{7.11}+\dfrac{9}{11.20}+\dfrac{5}{20.25}\)

Answer 1

a,\(A=\dfrac{1}{3.4}+\dfrac{1}{4.5}...+\dfrac{1}{29+30}\)

\(A=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{30}\)

\(A=\dfrac{1}{3}-\dfrac{1}{30}\)

\(A=\dfrac{9}{30}=\dfrac{3}{10}.\)

b, \(B=\dfrac{4}{7.11}+\dfrac{9}{11.20}+\dfrac{5}{20.25}\)

\(B=\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{25}\)

\(B=\dfrac{1}{7}-\dfrac{1}{25}\)

\(B=\dfrac{18}{175}\).