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28 tháng 3 2018

S = \(\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}+...+\frac{1}{17x19}\)

2S = \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\)\(\frac{1}{17}-\frac{1}{19}\)

2S = \(\frac{1}{3}-\frac{1}{19}\)

2S = \(\frac{16}{57}\)

S = \(\frac{16}{57}\times\frac{1}{2}\)

S = \(\frac{8}{57}\)

\(S=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}+\frac{1}{255}+\frac{1}{323}\)

\(S=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}\)

\(2S=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}\)

\(2S=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}\)

\(2S=\frac{1}{3}-\frac{1}{19}\)

\(2S=\frac{19}{57}-\frac{3}{57}\)

\(2S=\frac{16}{57}\)

\(S=\frac{16}{57}:2\)

\(S=\frac{16}{57}\cdot\frac{1}{2}\)

\(S=\frac{8}{57}\)

29 tháng 3 2018

Đặt \(A=1\frac{7}{15}-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}-\frac{1}{143}-\frac{1}{195}\)

\(\Rightarrow A=\frac{22}{15}-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\right)\)

Đặt \(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(\Rightarrow B=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(\Rightarrow2B=2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\right)\)

\(\Rightarrow2B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\)

\(\Rightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(\Rightarrow2B=1-\frac{1}{15}\)

\(\Rightarrow2B=\frac{14}{15}\)

\(\Rightarrow B=\frac{14}{15}:2\Rightarrow B=\frac{7}{15}\)

\(\Rightarrow A=\frac{22}{15}-\frac{7}{15}\Rightarrow A=\frac{15}{15}=1\)

29 tháng 3 2018

đáp án là 59​/15

   mình chắc chắn

                      

23 tháng 4 2018

2S=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

      = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\)

      =\(1-\frac{1}{15}=\frac{14}{15}\)

\(\Rightarrow S=\frac{7}{15}\)

23 tháng 4 2018

a. Ta có:A= 1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15

A=1/2(1/1.3+1/3.5+1/5.7+1/7.9+1/9.11+1/11.13+1/13.15)

A=1/2(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)

A=2(1-1/15)

A=1/2.14/15

A=7/15

26 tháng 7 2018


\(\frac{7}{15}\)nha ban

26 tháng 7 2018

bạn có thể trình bày cách làm cho mình ko

6 tháng 3 2016

a=8-\(\frac{8}{3.5}-\frac{8}{5.7}-\frac{8}{7.9}-\frac{8}{9.11}-\frac{8}{11.13}-\frac{8}{13.15}\)

a=8-\(\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}-\frac{1}{13}+\frac{1}{15}\)

a=8-1/3+1/15=126/15

14 tháng 2 2016

= 13/15 nha bn

duyệt đi

14 tháng 2 2016

A=1-1/15-1/35-1/63-1/99-1/143-1/195

=1-1/3.5-1/5.7-1/7.9-1/9.11-1/11.13-1/13.15

=1-1/2(2/3.5-2/5.7-2/7.9-2/9.11-2/11.13-2/13.15)

=1-1/2(1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+1/11-1/13+1/13-1/15)

=1-1/2.(1/3-1/15)

=1-1/2.4/15

=1-2/15=13/15

 

 

24 tháng 3 2019

a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\frac{14}{15}\)

\(=\frac{14}{30}=\frac{7}{15}\)

24 tháng 3 2019

a)

\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)

\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)

\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)

\(=2\left(1-\frac{1}{15}\right)\)

\(=2.\frac{14}{15}\)

\(=\frac{28}{15}\)

b)

\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)

                                                                                         \(...\)

10 tháng 1 2016

các bạn ghi ra câu trả lời rỏ ràng giùm mk nha.

22 tháng 3 2019

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

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

\(A=\frac{1}{2}\left(\frac{2}{3.5}+...+\frac{2}{11.13}\right)\)

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

\(A=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)\)

\(A=\frac{1}{2}\cdot\frac{10}{39}=\frac{5}{39}\)

P/s: Có thể tính sai :<

22 tháng 3 2019

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)

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

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

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

25 tháng 7 2018

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}\cdot\frac{14}{15}\)

\(=\frac{7}{15}\)

25 tháng 7 2018

Sửa đề chút nhé:

\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).0\)

\(=0\)

Ý b tham khảo bài bạn nguyen thi thuy linh nhé