\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}\text{ +}...\text{ +}\dfrac{1}{10.11.12}\)
K
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1
OY
26 tháng 9 2021
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}.\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
TA
9 tháng 4 2023
Ta có:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{10.11}-\dfrac{1}{11.12}=\dfrac{1}{1.2}-\dfrac{1}{11.12}=\dfrac{1}{2}-\dfrac{1}{132}=\dfrac{65}{132}\)Mà \(\dfrac{65}{132}\ne\dfrac{1}{4}\Rightarrow\) Có thể bạn ghi sai đề thì phải !
11 tháng 4 2023
\(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + .....+ \(\dfrac{1}{10.11.12}\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\) +....+ \(\dfrac{1}{10.11}\) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{1.2}\) + (- \(\dfrac{1}{2.3}\) + \(\dfrac{1}{2.3}\))+.......+ ( \(-\dfrac{1}{10.11}\) + \(\dfrac{1}{10.11}\)) - \(\dfrac{1}{11.12}\)
=\(\dfrac{1}{2}\) - \(\dfrac{1}{11.12}\) =\(\dfrac{1}{2}\) - \(\dfrac{1}{132}\) =\(\dfrac{66}{132}\)-\(\dfrac{1}{132}\) =\(\dfrac{65}{132}\) Vì \(\dfrac{33}{132}\) = \(\dfrac{1}{4}\) nên \(\dfrac{65}{132}\) > \(\dfrac{1}{4}\)
K
20 tháng 3 2018
Ta có :
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)
\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)
...
Do đó :
\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)
\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)
Vậy :
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)
TM
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:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
DL
29 tháng 3 2017
\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
29 tháng 3 2017
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...
TT
1
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
Đặt biểu thức trên là $A$.
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38.39}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{39-37}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)
\(\Rightarrow A=\frac{185}{741}\)
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:
\(M=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2004.2005.2006}\)
\(M=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{2004.2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2004.2005}-\dfrac{1}{2005.2006}\right)\)
\(M=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2005.2006}\right)\)
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{10.11.12}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{10.11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)
\(=\dfrac{1}{2}.\dfrac{65}{132}\)
\(=\dfrac{65}{264}\)
Vậy...