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30 tháng 8 2015

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=\frac{5}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{11\cdot13}\right)\)

\(=\frac{5}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{5}{2}\cdot\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}\cdot\frac{12}{13}\)

\(=\frac{30}{13}\)

30 tháng 8 2015

\(\frac{5}{3}+\frac{5}{15}+\frac{5}{35}+\frac{5}{63}+\frac{5}{99}+\frac{5}{143}\)

\(=5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{13}\right)\)

\(=\frac{5}{2}.\frac{12}{13}\)

\(=\frac{30}{13}\)

1: \(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\)

=1/2*10/39

=5/39

2: \(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\right)=\dfrac{5}{2}\cdot\dfrac{10}{11}=\dfrac{50}{22}=\dfrac{25}{11}\)

18 tháng 9 2023

\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

23 tháng 9 2018

\(B=\dfrac{4}{3}+\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{143}\) 

    \(=4(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143})\)

        vì \(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143}<\dfrac{1}{2}\) nên \(4(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{143})<4*\dfrac{1}{2}=2\Rightarrow B<2\)

25 tháng 7 2019

Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)

\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)

\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)

\(A=\frac{5}{39}\)

Câu còn lại cx dựa như vậy nhé bn ! 

Chúc bn hc tốt <3

25 tháng 7 2019

câu c hình như sai đề hả bn

15 tháng 4 2019

\(\frac{3x}{5}=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

Ta có: \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

      \(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)

      \(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)

      \(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)

      \(=\frac{1}{2}.\left(1-\frac{1}{13}\right)\)

      \(=\frac{1}{2}.\frac{12}{13}\)

      \(=\frac{6}{13}\)

\(\frac{3x}{5}=\frac{6}{13}\)

\(\Rightarrow3x=\frac{6.5}{13}\)

\(\Rightarrow3x=\frac{30}{13}\)

\(\Rightarrow x=\frac{10}{13}\)

~Học tốt~

22 tháng 4 2016

Ta có:

    A=5/15+5/35+5/63+5/99+...+5/2915

=>A=5/3.5+5/5.7+5/7.9+5/9.11+...+5/53.55

=>A=5/2.(2/3.5+2/5.7+2/7.9+2/9.11+...+2/53.55)

=>A=5/2.(2/3-2/5+2/5-2/7+2/7-2/9+2/9-2/11+...+2/53-2/55)

=>A=5/2.(2/3-2/55)

=>A=5/2.104/165

=>A=52/33

Vậy A=52/33

OK!

23 tháng 4 2016

thank