Question

Tính tổng

A=1/1.2+1/2.3+.......+1/49/50

B=2/3.5+2/5.7+..........+2/37.39

C=3/4.7+3/7.10+.........+3/37.39

Ai giúp mình với!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{49}{50}\)

Answer 2

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\dfrac{1}{1}-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}\)

Answer 3

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)

= \(\left(1-\dfrac{1}{2}\right)\)+\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\)+...+\(\left(\dfrac{1}{49}-\dfrac{1}{50}\right)\)

= \(\left(1-\dfrac{1}{50}\right)\) = \(\dfrac{49}{50}\)

Answer 4

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(B=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{12}{39}-\dfrac{1}{39}=\dfrac{11}{39}\)

Answer 5

\(B=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)

\(B=\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{37.39}\right)\)

\(B=\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\)

\(B=\left(\dfrac{1}{3}-\dfrac{1}{39}\right)=\dfrac{4}{13}\)

\(C=\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.39}\)

\(C=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{39}\)

\(C=\dfrac{1}{4}-\dfrac{1}{39}=\dfrac{35}{156}\)

Answer 6

\(C=\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{36.39}=\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{36}-\dfrac{1}{39}\)

\(C=\dfrac{1}{4}-\dfrac{1}{36}=\dfrac{9}{36}-\dfrac{1}{36}=\dfrac{8}{36}=\dfrac{4}{18}\)

Answer 7

A = \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{49}-\dfrac{1}{50}\)

= 1 - \(\dfrac{1}{50}\) = \(\dfrac{49}{50}\)

Vậy A = \(\dfrac{49}{50}\)

B = \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+.....+\dfrac{2}{37.39}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{37}-\dfrac{1}{39}\)

= \(\dfrac{1}{3}-\dfrac{1}{39}\) = \(\dfrac{13}{39}-\dfrac{1}{39}=\dfrac{38}{39}\)

Vậy B = \(\dfrac{38}{39}\)

Answer 8

C = \(\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{37.39}\)

= \(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{37}-\dfrac{1}{39}\)

= \(\dfrac{1}{4}-\dfrac{1}{39}\) = \(\dfrac{35}{156}\)

Vậy C = \(\dfrac{35}{156}\)