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12 tháng 10 2018

14 tháng 6 2023

a) \(A=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2003}\right).\left(1-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{2002}{2003}.\dfrac{2003}{2004}\)

\(=\dfrac{1}{2004}\)

b) \(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)

\(=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right).\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{35}{6}.\dfrac{4}{7}\)

\(=\dfrac{59}{15}-\dfrac{10}{3}\)

\(=\dfrac{3}{5}\)

14 tháng 6 2023

ai trả lời đúng thì mình sẽ tick cho

16 tháng 7 2021

giúp mình vớiiii

 

21 tháng 12 2022

`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`

`=1/2xx2/3xx3/4xx4/5`

`=[1xx2xx3xx4]/[2xx3xx4xx5]`

`=1/5`

`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`

`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`

`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`

`=1/100`

8 tháng 5 2022

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}\)

8 tháng 5 2022

ai đó giúp mik vs :((

5 tháng 6 2023

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

\(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

\(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

\(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)

20 tháng 8 2018

a) \(\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

                                                                                      \(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

                                                                                     \(=\frac{1}{2}-\frac{1}{6}=\frac{3}{6}-\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\)

b) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}=\frac{1}{4}\)

20 tháng 8 2018

\(a)\frac{1}{2}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}+\frac{1}{4}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{6}\)

\(\Rightarrow\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow\frac{1}{3}\)

\(b)\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\)

\(=\frac{1\times2\times3}{2\times3\times4}\)

\(=\frac{1}{4}\)

\(b)\left(1-\frac{1}{2}\right)\div\left(1-\frac{1}{3}\right)\div\left(1-\frac{1}{4}\right)\)

\(=\frac{1}{2}\div\frac{2}{3}\div\frac{3}{4}\)

\(=\frac{1}{2}\times\frac{3}{2}\times\frac{4}{3}\)

\(=\frac{1\times3\times2\times2}{2\times2\times3}\)

\(=1\)