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21 tháng 6 2020

Đề bài : Tính

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{20}\right)\)

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

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{19}{20}\)

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

21 tháng 6 2020

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{19}{20}\)

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

26 tháng 8 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)

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

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

\(=\frac{4}{5}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

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

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

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

26 tháng 8 2016

Từng ý một nhanh hơn nhá

21 tháng 9 2024

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

   

21 tháng 9 2024

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

 

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{20}\right)=\frac{1}{2}.\frac{2}{3}...\frac{19}{20}=\frac{1.2...19}{2.3...20}=\frac{1}{20}\)

4 tháng 8 2016

a) 4/5(3/7+4/7)=4/5

câu b,c cũng dễ

4 tháng 8 2016

a) 4/5 x 3/7 + 4/5 x 4/7

= 4/5 x ( 3/7 + 4/7 )

= 4/5 x 1

= 4/5

29 tháng 6 2016

giup minh voi

29 tháng 6 2016

1/1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20

=1/1-1/20=19/20

\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)

\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)

\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)

\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)

\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)

\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)

\(=52-\frac{246}{7}\div\frac{82}{63}\)

\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)

\(=52-27=25\)

\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)

\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)

\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)

\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)

\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)

\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)

\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)

21 tháng 6 2020

\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{19}{20}\)

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

21 tháng 6 2020

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot....\cdot\frac{19}{20}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}=\frac{1}{20}\)

3 tháng 11 2018

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)(Dùng cộng rồi trừ chính số đó bằng 0)

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

=\(\frac{10}{20}-\frac{1}{20}\)( Dùng phương pháp quy đồng)

=\(\frac{9}{20}\)