Question

Tính nhanh:

A=\(\dfrac{1}{2.3}\) +\(\dfrac{1}{3.4}\) +\(\dfrac{1}{4.5}\) +....+\(\dfrac{1}{99.100}\).

B=\(\dfrac{3}{2.5}\) +\(\dfrac{3}{5.8}\)+\(\dfrac{3}{8.11}\)+....\(\dfrac{3}{92.95}\).

C=\(\dfrac{7}{1.3}\) +\(\dfrac{7}{3.5}\)+\(\dfrac{7}{5.7}\)+\(\dfrac{7}{7.9}\)+.....+\(\dfrac{7}{49.51}\)

Answer 1

A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)

Answer 2

B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)

B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)

Answer 3

C = \(\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{49.51}\)

C = \(\dfrac{2}{2}.\left(\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{49.51}\right)\)

C = \(\dfrac{7}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{49.51}\right)\)

C = \(\dfrac{7}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

C = \(\dfrac{7}{2}.\left(1-\dfrac{1}{51}\right)\)

C =\(\dfrac{7}{2}.\dfrac{49}{51}\)

C =\(\dfrac{343}{102}\)

Answer 4

Công thức

\(\dfrac{a-b}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}\)

Answer 5

A= 1/2-1/3+1/3-1/4+.....+1/99-1/100

A=1/2-1/100=49/100

B=1/2-1/5+1/5-1/8+....+1/92-1/95

B=1/2-1/95=93/190

C=7/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/49-1/51)

C=7/2.(1-1/51)

C=7/2.50/51=175/51

Answer 6

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)

=> \(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

=> \(A=\dfrac{1}{2}-\dfrac{1}{100}\)

=> \(A=\dfrac{49}{100}\)

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\(B=\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{92.95}\)

=> \(B=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}\)

=> \(B=\dfrac{1}{2}-\dfrac{1}{95}\)

=> \(B=\dfrac{93}{190}\)

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\(C=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{49.51}\)

=> \(2C=\dfrac{7.2}{1.3}+\dfrac{7.2}{3.5}+\dfrac{7.2}{3.7}+\dfrac{7.2}{7.9}+...+\dfrac{7.2}{49.51}\)

=> \(2C=7.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{49.51}\right)\)

=> \(2C=7.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

=> \(2C=7.\left(1-\dfrac{1}{51}\right)\)

=> \(2C=7.\dfrac{50}{51}\)

=> \(2C=\dfrac{350}{51}\)

=> \(C=\dfrac{350}{51}:2\)

=> \(C=\dfrac{175}{51}\)