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30 tháng 1 2016

\(\frac{1}{1.3}+\frac{1}{3.2}+\frac{1}{2.5}+...+\frac{1}{99.100}\)

\(2.\left(\frac{1}{1.3.2}+\frac{1}{3.2.2}+\frac{1}{2.5.2}+...+\frac{1}{99.50.2}\right)\)

\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(2.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(2.\frac{49}{100}\)

\(\frac{49}{50}\)

8 tháng 8 2023

a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)

Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)

\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)

\(\dfrac{1}{x+1}=\dfrac{1}{100}\)

\(\Rightarrow x+1=100\)

\(x=100-1\)

\(x=99\)

8 tháng 8 2023

câu b thiếu kết quả đúng không bn?

22 tháng 4 2023

=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2 

=(1-1/9)chia 2

=8/9 chia 2

=4/9

24 tháng 4 2023

Đặt �=11�3+13�5+15�7+17�9

2�=21�3+23�5+25�7+27�9

2�=11−13+13−15+...+17−19

2�=11−19=89

�=89.12=49
 

15 tháng 12 2023

\(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+...+\dfrac{1}{121\times123}\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{121}-\dfrac{1}{123}\right)\)

\(=\dfrac{1}{2}.\left(1-\dfrac{1}{123}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{123}{123}-\dfrac{1}{123}\right)\)

\(=\dfrac{1}{2}.\dfrac{122}{123}\)

\(=\dfrac{61}{123}\)

\(#NqHahh\)

13 tháng 8 2016

Đặt \(A=\frac{1}{1x3}+\frac{1}{3x5}+\frac{1}{5x7}+\frac{1}{7x9}\)

\(2A=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}\)

\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\)

\(2A=\frac{1}{1}-\frac{1}{9}=\frac{8}{9}\)

\(A=\frac{8}{9}.\frac{1}{2}=\frac{4}{9}\)

13 tháng 8 2016

A= 1/(1x3) + 1/(3x5)+ 1/(5x7) + 1/(7x9) +  1/(9x11)

A x 2 = 2/(1x3) + 2/(3x5)+ 2/(5x7) + 2/(7x9) +  2/(9x11)

Nhận xét :

2/(1x3) = 1 - 1/3

2/(3x5) = 1/3 - 1/5

2/(5x7) = 1/5 - 1/7

2/(7x9) = 1/7 - 1/9

2/(9x11) = 1/9 - 1/11

A x 2 = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

A x 2 = 1 - 1/11

A x 2 = 10/11

A = 10/11 : 2 = 5/11

các bạn k mình nha!

7 tháng 5 2015

Coi A=1/1x3+1/3x5+1/5x7+1/7x9

=>2A=2x(1/1x3+1/3x5+1/5x7+1/7x9)=2/1x3+2/3x5+2/5x7+2/7x9

=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

=1-1/9=8/9

=>A=8/9:2=4/9

9 tháng 4 2015

\(\frac{1}{1x3}+\frac{1}{3x5}+....+\frac{1}{97x99}\)=S

 

\(2S=\frac{3-1}{1x3}+\frac{5-3}{3x5}+...+\frac{99-97}{97x99}\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

\(S=\frac{2S}{2}=\frac{49}{99}\)

1/1 x 3 + 1/3 x 5 + 1/5 x 7 + 1/7 x 9 + 1/9 x 11

= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11

= 1 - 1/11

= 10/11