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\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)x=1\)

\(\frac{1}{2}\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)=\frac{1}{x}\)

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

\(\frac{1}{3}-\frac{1}{11}=\frac{2}{x}\)

\(\frac{2}{x}=\frac{8}{33}\)

\(8x=66\)

\(x=66:8=\frac{33}{4}\)

29 tháng 7 2019

\(\left(\frac{1}{15}+\frac{1}{25}+\frac{1}{63}+\frac{1}{99}\right).x=1\)

\(\Leftrightarrow\frac{766}{5775}.x=1\)

\(\Leftrightarrow x=1:\frac{766}{5775}\)

\(\Leftrightarrow x=\frac{5775}{766}\)

Hông chắc nhé bn !!!

29 tháng 7 2019

\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\times x=1\)

\(\Rightarrow\left(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}\right)\times x=1\)

\(\Rightarrow\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times x=1\)

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

\(\Rightarrow\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{11}\right)\times x=1\)

\(\Rightarrow\frac{1}{2}\times\frac{8}{33}\times x=1\)

\(\Rightarrow\frac{4}{33}\times x=1\)

\(\Rightarrow x=1:\frac{4}{33}\)

\(\Rightarrow x=\frac{33}{4}\)

Vậy \(x=\frac{33}{4}\)

23 tháng 8 2023

\(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)

\(=\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{11}-\dfrac{1}{13}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\left(1-\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{2}{3}\cdot\dfrac{1}{2}+\dfrac{1}{25}\)

\(=\dfrac{1}{3}+\dfrac{1}{25}\)

\(=\dfrac{28}{75}\)

24 tháng 8 2023

cám ơi bạn

 

1 tháng 7 2019

\(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+.......\frac{1}{13x15}=\frac{1}{2}x\frac{2}{1x3}+\frac{2}{3x5}.......+\frac{2}{13x15}\)

\(A=\frac{1}{2}x\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\right)\)

Còn lại em nhân giống ở trên nhé

1 tháng 7 2019

Đặt A = 1/15 + 1/35 + ... + 1/3135 

       A = 1/3.5 + 1/5.7 + ... + 1/55.57

     2A =  2/3.5 + 2/5.7 + ... + 2/55.57 

    2A = 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/55 - 1/57 

    2A = 1/3 - 1/57 = 6/19 

      A = 3/19 

30 tháng 3 2019

\((\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99})x=\frac{2}{3}\)

Đặt \(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

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

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

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

\(A=\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(A=\frac{1}{2}.\frac{10}{11}=\frac{5}{11}\)

Thay A vào biểu thức

\(\Rightarrow\frac{5}{11}x=\frac{2}{3}\)

\(\Rightarrow x=\frac{22}{15}\)

P/s: Có thể tính sai :(

30 tháng 3 2019

\(\left[\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right]\times x=\frac{2}{3}\)

Trước tiên mình tính dãy có dấu ngoặc đã

Đặt : \(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\)

\(=\frac{1}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{11}\right]\)

\(=\frac{1}{2}\left[1-\frac{1}{11}\right]=\frac{1}{2}\cdot\frac{10}{11}=\frac{1\cdot10}{2\cdot11}=\frac{1\cdot5}{1\cdot11}=\frac{5}{11}\)

Thay vào biểu thức \(S=\frac{5}{11}\)ta lại có :

\(\frac{5}{11}\times x=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{2}{3}:\frac{5}{11}\)

\(\Leftrightarrow x=\frac{2}{3}\cdot\frac{11}{5}\)

\(\Leftrightarrow x=\frac{22}{15}\)

Vậy \(x=\frac{22}{15}\)

18 tháng 8 2017

<=> \(\left(\frac{1}{3\cdot5}+\frac{1}{5.7}+...+\frac{1}{13\cdot15}\right)+x=\frac{17}{15}\)

<=> \(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=>\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)+x=\frac{17}{15}\)

<=> \(\frac{2}{15}+x=\frac{17}{15}\)

=> x = 1

18 tháng 8 2017

(1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)+x=17/15

[2.(1/3-1/5+1/5-1/7+...+1/13-1/15)]+x=17/15

[2.(1/3-1/15)]+x=17/15

(2.4/15)+x=17/15

6/15+x=17/15

x=17/15-6/15

x=11/15

3 tháng 7 2017

=> 2(1/15+1/35+1/63+1/99)x=2

=>(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11)x=2

=>8/33x=2

=>x=2:8/33

=>x=8,25

3 tháng 7 2017

\(\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\cdot x=1\)

\(\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}\right)\cdot x=1\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]\cdot x=1\)

\(\left[\frac{1}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\right]\cdot x=1\)

\(\left[\frac{1}{2}\cdot\frac{8}{33}\right]\cdot x=1\)

\(\frac{4}{33}\cdot x=1\)

\(\Rightarrow x=\frac{1}{\frac{4}{33}}=\frac{33}{4}\)

Đặt phép tính cần tìm là A

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

\(2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(2A=1-\dfrac{1}{13}\)

\(2A=\dfrac{12}{13}\)

\(A=\dfrac{6}{13}\)

HQ
Hà Quang Minh
Giáo viên
21 tháng 8 2023

\(A=\dfrac{1}{3}+\dfrac{1}{15}+...+\dfrac{1}{143}\\ =\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\\ =\dfrac{1}{2}\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{11\times13}\right)\\ =\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\right)\\ =\dfrac{1}{2}\times\dfrac{12}{13}\\ =\dfrac{6}{13}\)