Đặt  \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)

\(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)

\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(A=2\left(1-\frac{1}{20}\right)\)

\(A=2.\frac{19}{20}=\frac{19}{10}\)

Vậy ...

=2.(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{19.20}\))

=2.( 1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..........+\(\frac{1}{19}\)-\(\frac{1}{20}\))

=2.(1-\(\frac{1}{20}\))

=2.\(\frac{19}{20}\)

=  \(\frac{19}{10}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2.\left(1-\frac{1}{20}\right)\)

\(=2\cdot\frac{19}{20}=\frac{19}{10}\)

Đặt tổng trên là A , ta có :

\(\frac{A}{2}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(\frac{A}{2}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)\)

\(\frac{A}{2}=\left(1-\frac{1}{100}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{98}\right)+\left(\frac{1}{99}-\frac{1}{99}\right)\)\(\frac{A}{2}=\frac{99}{100}\)

\(A=\frac{99}{100}.2\)

\(A=\frac{99}{50}\)

1/6 nhe

\(=\frac{1}{6}\)

\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}.\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot\frac{5^2}{5\cdot6}\)

\(=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6}\)

\(=\frac{1}{6}\)

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

\(S=\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{98\times99}+\frac{2}{99\times100}\)

\(S=2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(S=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(S=2\times\left(1-\frac{1}{100}\right)\)

\(S=2\times\frac{99}{100}\)

\(S=\frac{99}{50}\)

\(S=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{98.99}+\frac{2}{99.100}\)

\(S=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}+\frac{1}{100}\right)\)

\(S=2.\left(\frac{1}{1}-\frac{1}{100}\right)\\ S=2.\left(\frac{100}{100}+\frac{-1}{100}\right)\\ S=2.\frac{99}{100}\\ S=\frac{99}{50}\)

/hoi-dap/question/115788.html Giúp mik vs\

dấu bằng của mk bt liệt nên bạn thông cảm 

A bằng (1.2.3.4).(1.2.3.4)/(1.2.3.4).(2.3.4.5) bằng 5 

rút gọn cho nhau bạn nhé

thank bạn nhóa

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

=1-1/20

=19/20

Đặt biểu thức trên là A

Ta có: A =(1^2 . 2^2 . 3^2 . 4^2)/(1.2.2.3.3.4.4.5)

              = [(1.2.3.4).(1.2.3.4)] / [(1.2.3.4).(2.3.4.5)]

               = 1/5

Vậy A = 1/5

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)

=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)

=\(\frac{1}{5}\)

sorry mình nhầm

ta có:

M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)

=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)

=\(\frac{1}{5}\)

vậy M=\(\frac{1}{5}\)

\(M=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

ta có:

\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).