S= 1x2+2x3+3x4+4x5+...+ 20x21

3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )

3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3

3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)

3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21

3xS = 20x21x22

 S = 20x21x22 /3

S= 1x2+2x3+3x4+4x5+...+ 20x21

3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )

3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3

3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)

3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21

3xS = 20x21x22

S = 20x21x22 /3

k mk nha

S= 1x2+2x3+3x4+4x5+...+ 20x21 3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 ) 3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3 3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19) 3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21 3xS = 20x21x22 S = 20x21x22 /3

ta có\(A=\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)

             \(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\right)\)

             \(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)

               \(=4\left(1-\frac{1}{2015}\right)\)

                \(=4\cdot\frac{2014}{2015}\)

                  \(=\frac{8056}{2015}\)

  VẬY A=\(\frac{8056}{2015}\)

Ta có:\(A=\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

\(=9\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=9\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=9\left(1-\frac{1}{100}\right)\)

\(=9.\frac{99}{100}=\frac{891}{100}\)

A = \(\frac{1}{1\times2}\) + \(\frac{1}{2\times3}\) + \(\frac{1}{3\times4}\) + ... + \(\frac{1}{19\times20}\) + \(\frac{1}{20\times21}\)

A = \(\frac11\) - \(\frac12\) + \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + ... + \(\frac{1}{19}\) - \(\frac{1}{20}\) + \(\frac{1}{20}\) - \(\frac{1}{21}\)

A = \(\frac11\) - \(\frac{1}{21}\)

A = \(\frac{20}{21}\)

Ta có: \(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{20\times21}\)

\(=1-\frac12+\frac12-\frac13+\cdots+\frac{1}{20}-\frac{1}{21}\)

\(=1-\frac{1}{21}=\frac{20}{21}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\ldots+\frac{1}{19\times20}+\frac{1}{20\times21}\)

\(=\frac11-\frac12+\frac12-\frac13+\frac14-\frac14+\ldots+\frac{1}{20}-\frac{1}{21}\)

\(=1-\frac{1}{21}\)

Vậy kết quả của phép tính trên là \(=\frac{20}{21}\)

lam ơn trả lời câu này

\(\frac{2019}{1\times2}+\frac{2019}{2\times3}+\frac{2019}{3\times4}+...+\frac{2019}{2018\times2019}\)

\(=2019\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2018\times2019}\right)\)

\(=2019\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)

\(=2019\left(1-\frac{1}{2019}\right)\)

\(=2019\left(\frac{2019}{2019}-\frac{1}{2019}\right)\)

\(=2019\times\frac{2018}{2019}\)\(=\frac{2019\times2018}{2019}=2018\)