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2 tháng 2 2023

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+2.\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+2.\left(\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2005.2006}\right)\)

\(=1-\dfrac{2}{2005.2006}\)

\(=\dfrac{2011014}{2011015}\).

Ta có:

\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)

\(M=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2004.2005.2006}\right)\)

\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)

\(M=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2005.2006}\right)\)

 

AH
Akai Haruma
Giáo viên
31 tháng 7 2021

Lời giải:

Đặt biểu thức trên là $A$.
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)

\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)

\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)

\(\Rightarrow A=\frac{185}{741}\)

 

 

1 tháng 10 2023

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{48\cdot49\cdot50}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-...+\dfrac{1}{48\cdot49}-\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{1\cdot2}-\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2450}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{612}{1225}\)

\(=\dfrac{306}{1225}\)

AH
Akai Haruma
Giáo viên
31 tháng 5 2023

Lời giải:

Gọi tổng trong ngoặc là $A$
$2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{10-8}{8.9.10}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}$

$=\frac{1}{1.2}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{90}=\frac{22}{45}$

Vậy $\frac{22}{45}x=\frac{23}{45}$

$\Rightarrow x=\frac{23}{45}: \frac{22}{45}=\frac{23}{22}$

AH
Akai Haruma
Giáo viên
31 tháng 3 2023

$x$ ở cuối là sao đây bạn? Nhân riêng với $\frac{1}{8.9.10}$ à?

14 tháng 5 2023

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{37.38.39}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)

\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{38.39}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1482}\right)\)

\(A=\dfrac{1}{2}.\dfrac{370}{741}=\dfrac{185}{741}\)

26 tháng 9 2021

\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)

\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

17 tháng 5 2022

\(2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{98.99.100}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{50.99-1}{100.99}=\dfrac{4949}{9900}\)

17 tháng 5 2022

`A=1/[1.2.3]+1/[2.3.4]+....+1/[98.99.100]`

`A=1/2.(2/[1.2.3]+2/[2.3.4]+....+2/[98.99.100])`

`A=1/2.(1/[1.2]-1/[2.3]+1/[2.3]-1/[3.4]+....+1/[98.99]-1/[99.100])`

`A=1/2.(1/[1.2]-1/[99.100])`

`A=1/2.(1/2-1/9900)`

`A=1/2.(4950/9900-1/9900)`

`A=1/2 . 4949/9900`

`A=4949/19800`