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1 tháng 9 2017

A= \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{4.5.6}+....+\dfrac{1}{37.38.39}\)

A=\(\dfrac{1}{1}-\dfrac{1}{39}\)

A=\(\dfrac{38}{39}\)

còn lại tự làm do mình có việc chút

31 tháng 8 2017

Chưa học

26 tháng 4 2017

a) Ta có:

3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)

A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)

Lấy (1) - (2) ta được:

1-\(\dfrac{1}{3^{100}}\)

b) Ta xét:

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)

Ta có:

2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)

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

=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)

Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)

=\(\dfrac{370}{741}\)

Nếu bn cảm thấy mk đúng tick cho mk nhé!

haha

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

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

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

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

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

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

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

AH
Akai Haruma
Giáo viên
31 tháng 7 2021

Lời giải:

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

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

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

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

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

 

 

16 tháng 3 2021

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

22 tháng 4 2021

Tìm y:

-y:1/2-5/2=4+1/2

-y:1/2 = 4+1/2+5/2

-y:1/2 = 7

-y = 7.2

y = -14

Vậy y = -14

5 tháng 2 2018

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

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

\(=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)

\(=\dfrac{370}{741}\)

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

Ta có:

\(\dfrac{1}{1.2.3}=\dfrac{1}{1.2}-\dfrac{1}{2.3}\); \(\dfrac{1}{2.3.4}=\dfrac{1}{2.3}-\dfrac{1}{3.4}\);.......

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

\(\Rightarrow A=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)

\(=\dfrac{370}{741}\)

Vậy \(A=\dfrac{370}{741}\)

1 tháng 12 2018

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

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

\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

Vậy..

NV
1 tháng 12 2018

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

\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)