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DT
25 tháng 6 2023

`a)` Xét tử số phân số M :

\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)

Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)

 

DT
25 tháng 6 2023

`b)` Xét tử số phân số N :

\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

Xét mẫu số phân số N :

\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)

\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)

Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)

AH
Akai Haruma
Giáo viên
20 tháng 9 2021

Lời giải:
\(299A=\frac{300-1}{1.300}+\frac{301-2}{2.301}+\frac{302-3}{3.302}+....+\frac{400-101}{101.400}\)

\(=1-\frac{1}{300}+\frac{1}{2}-\frac{1}{301}+\frac{1}{3}-\frac{1}{302}+...+\frac{1}{101}-\frac{1}{400}\)

\(=(1+\frac{1}{2}+....+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(1)\)

Mặt khác:

$101B=\frac{102-1}{1.102}+\frac{103-2}{2.103}+...+\frac{400-299}{299.400}$

$=1-\frac{1}{102}+\frac{1}{2}-\frac{1}{103}+....+\frac{1}{299}-\frac{1}{400}$

$=(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{299})-(\frac{1}{102}+\frac{1}{103}+....+\frac{1}{400})$

$=(1+\frac{1}{2}+...+\frac{1}{101})-(\frac{1}{300}+\frac{1}{301}+...+\frac{1}{400})(2)$

Từ $(1);(2)\Rightarrow 299A=101B$

$\Rightarrow \frac{A}{B}=\frac{101}{299}$

21 tháng 3 2023

sai r

 

12 tháng 12 2021

S = \(\left(1+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2020}\right)\)

\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1010}\right)\)

\(\dfrac{1}{1011}+\dfrac{1}{1012}+...+\dfrac{1}{2021}\)

30 tháng 4

S=P nhé

 

27 tháng 2 2019

\(M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\right):\dfrac{2012}{2013}\)

\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right).\dfrac{2013}{2012}\)

\(M=0.\dfrac{2013}{2012}\)

\(M=0\)

1 tháng 12 2017

Chữa lại đề.Bạn xem lại đề xem đúng chưa nhé!

\(D=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}+\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}+\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}+\dfrac{3}{2004}}\)

\(D=\dfrac{1.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}{5.\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}\right)}-\dfrac{2.\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}{3\left(\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)}\)

\(D=\dfrac{1}{5}-\dfrac{2}{3}\)

\(D=-\dfrac{7}{15}\)

Cái này học lâu rồi.Bạn xem lại xem mình làm đúng chưa nhé!

1 tháng 12 2017

làm H đi tui cx đang cằn

16 tháng 6 2018

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

=> x+1=0

<=> x=-1

b) \(\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)

<=> \(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

đến đây tương tự a

16 tháng 6 2018

a) Ta có:

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\left(Vì:\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)

\(\Leftrightarrow x=-1\)

Vậy....

b)Sửa lại đề nha

Ta có:

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)

\(\Leftrightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)

Lý giải tương tự câu a và kết luận nha

2 tháng 3 2017

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}\)

\(=\frac{1}{2014}\)

Vậy \(A=\frac{1}{2014}\)

2 tháng 3 2017

Đặt B=\(2012+\dfrac{2012}{2}+\dfrac{2011}{3}+...+\dfrac{1}{2013}\)

=>B=\(\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{1}{2013}\right)\)

=\(\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}\)

=\(2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)\)

=>A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2013}}{2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2013}\right)}=\dfrac{1}{2014}\)

Vậy ...

28 tháng 2 2023

Nhận xét nè: ở mẫu số tại các phân số có tử số + mẫu số = 2012. Vì vậy mục tiêu là tạo ra con 2012 ở các phân số của mẫu số. E xử con tử số ở phân số mẫu số sao cho ra con 2012 là được =))

6 tháng 8 2017

\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)

\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)

\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)

\(\Leftrightarrow x=2014\)

Vậy \(x=2014\)

6 tháng 8 2017

\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)

\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)

Vậy x = 2014