Question

\(C=\left(\dfrac{4}{1\cdot3}\right)+\left(\dfrac{4}{3\cdot5}\right)+...+\left(\dfrac{4}{97\cdot99}\right)-\dfrac{5}{11\cdot12}-\dfrac{5}{12\cdot13}-...-\dfrac{5}{98\cdot99}\)

ok, giải nhanh cho mk nha

Answer 1

Xét C = \(\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\right)\)

Đặt A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)

B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)

=> C = A - B

Ta có : A = \(\dfrac{4}{1.3}+\dfrac{4}{3.5}+...+\dfrac{4}{97.99}\)

= 2 \(\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)\)

= \(2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)

= \(2\left(1-\dfrac{1}{99}\right)=\dfrac{2.98}{99}=\dfrac{196}{99}\)

Ta có B = \(\dfrac{5}{11.12}+\dfrac{5}{12.13}+...+\dfrac{5}{98.99}\)

= \(5\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{98.99}\right)\)

= \(5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)=\dfrac{8.5}{99}=\dfrac{40}{99}\)

=> C = A - B = \(\dfrac{196-40}{99}=\dfrac{156}{99}=\dfrac{52}{33}\)

Answer 2

\(C=\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}-\dfrac{5}{11.12}-\dfrac{5}{12.13}-.....-\dfrac{5}{98.99}\)
\(C=\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+.....+\dfrac{4}{97.99}\right)-\left(\dfrac{5}{11.12}+\dfrac{5}{12.13}+.....+\dfrac{5}{98.99}\right)\)\(C=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+.....+\dfrac{1}{97}-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+.....+\dfrac{1}{98}-\dfrac{1}{99}\right)\)\(C=2\left(1-\dfrac{1}{99}\right)-5\left(\dfrac{1}{11}-\dfrac{1}{99}\right)\)

\(C=2.\dfrac{98}{99}-5.\dfrac{8}{99}\)

\(C=\dfrac{196}{99}-\dfrac{40}{99}=\dfrac{52}{33}\)