\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.....+\dfrac{1}{37.38.39}\)
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S
1
NA
Nguyễn An Ninh
VIP
14 tháng 5 2023
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{37.38.39}\)
\(A=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{37.38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{38.39}\right)\)
\(A=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{1482}\right)\)
\(A=\dfrac{1}{2}.\dfrac{370}{741}=\dfrac{185}{741}\)
SC
2
5 tháng 2 2018
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.......+\dfrac{1}{37.38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.....+\dfrac{1}{37.38}-\dfrac{1}{38.39}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{38.39}\)
\(=\dfrac{370}{741}\)
5 tháng 2 2018
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+......+\dfrac{1}{37.38.39}\)
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 \(A=\dfrac{370}{741}\)
PT
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!
AN
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
GL
2
YN
2 tháng 2 2023
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+2.\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+2.\left(\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(=2.\left(\dfrac{1}{1.2}-\dfrac{1}{2005.2006}\right)\)
\(=1-\dfrac{2}{2005.2006}\)
\(=\dfrac{2011014}{2011015}\).
2 tháng 2 2023
Ta có:
\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(M=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2004.2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2005.2006}\right)\)
AN
1
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?
Lời giải:
Đặt biểu thức trên là $A$.
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)
\(\Rightarrow A=\frac{185}{741}\)