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2 tháng 6 2015

A=3/2(2/1.3+2/3.5+2/5.7+....+2/53.55)

=3/2(1-1/3+1/3-1/5+1/5-1/7+..../1/53-1/55)

=3/2(1-1/55)

=3/2.54/55

=81/55

11 tháng 5 2017

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{50}{51}\)

11 tháng 5 2017

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(2A=3\left(1-\frac{1}{51}\right)\)

\(2A=3.\frac{50}{51}\)

\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'

4 tháng 8 2016

\(C=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

4 tháng 8 2016

\(C=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)

\(C=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(C=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)

5 tháng 4 2016

=3(1/1.3+1/3.5+1/5.7+1/7.9)

=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2

=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0

=3/2.8/9=4/3

5 tháng 4 2016

ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)

ta đặt 3 làm chung rồi tự làm đc

16 tháng 5 2016

A=3/1*3+3/3*5+3/5*7+...+3/2015*2017

A=3/2*(2/1*3+2/3*5+2/5*7+...+2/2015*2017)

A=3/2*(1-1/3+1/3-1/5+1/5-1/7+...+1/2015-1/2017)

A=3/2*(1-1/2017)

A=3/2*2016/2017

A=3024/2017

16 tháng 5 2016

A= \(\frac{3}{1.3}\)+\(\frac{3}{3.5}\)+\(\frac{3}{5.7}\)+....+\(\frac{3}{2015.2017}\)

A= \(\frac{3}{2}\).(\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+...+\(\frac{2}{2015.2017}\))

A= \(\frac{3}{2}\).( 1- \(\frac{1}{3}\)\(\frac{1}{3}\)\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+... \(\frac{1}{2015}\)\(\frac{1}{2017}\))

A= \(\frac{3}{2}\).(1- \(\frac{1}{2017}\))

A= \(\frac{3}{2}\)\(\frac{2016}{2017}\)

A= \(\frac{3024}{2017}\)

25 tháng 4 2018

Ta có : 

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

Vậy \(A=\frac{25}{17}\)

Chúc bạn học tốt ~ 

25 tháng 4 2018

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)

\(A=\frac{3}{2}.\frac{50}{51}\)

\(A=\frac{25}{17}\)

\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(B=\frac{21}{4}.33.\frac{4}{21}\)

\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)

\(B=33\)

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)

\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(C=\frac{1}{2}.\frac{98}{99}\)

\(C=\frac{49}{99}\)

7 tháng 4 2017

k mình nha

4 tháng 5 2016

 nhung ma ko cothoi gian giai

4 tháng 5 2016

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

2 tháng 4 2016

A=3/1.3+3/3.5+3/5.7+............+3/49.51

A=3/1-3/3=3/3-3/5+3/5-3/7+...............+3/49-3/51

A=1-1/3+1/3-1/5+1/5-1/7+.....................+1/39-1/51

A=1-1/51

A=50/51

2 tháng 4 2016

A\(=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{49.51}\right) \)

    \(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...\frac{2}{49.51}\right)\)

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

     =\(\frac{3}{2}\left(1-\frac{1}{51}\right)\) 

    \(=\frac{3}{2}.\frac{50}{51}\)   

  \(=\frac{25}{17}\)