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Đặt B= \(\dfrac{2011}{1}+\dfrac{2010}{2}+.......+\dfrac{1}{2011}\)
Cộng 1 vào ta được:
B=(\(\dfrac{2012}{1}+\dfrac{2012}{2}+.......+\dfrac{2012}{2011}\)+\(\dfrac{2012}{2012}\)) -2012
-> B= 2012 (\(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2011}\)+\(\dfrac{1}{2012}\)) -2012+\(\dfrac{2012}{1}\)
Thay vào P ta được:
P=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2012}}{2012\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2012}\right)}\)
-> P= \(\dfrac{1}{2012}\)
có chỗ nào chưa hiểu hỏi mình nha!
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!
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}\)
=>3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}\)
=>3B-B=2B=1-\(\dfrac{1}{3^{2012}}\)
=>B=\(\dfrac{1}{2}-\dfrac{1}{2.3^{20112}}\)<1/2
vậy........
\(D=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)
Ta có mẫu của phân số trên là :
\(\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}\)
\(=\left(\dfrac{2010}{2}+1\right)+\left(\dfrac{2009}{3}+1\right)+...+\left(\dfrac{1}{2011}+1\right)+1\)
=\(\dfrac{2012}{2}+\dfrac{2012}{3}+\dfrac{2012}{4}+...+\dfrac{2012}{2011}+\dfrac{2012}{2012}\)
=\(2012\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2012}\right)\)
Từ đó suy ra :
\(D=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{2012\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)}=\dfrac{1}{2012}\)
Vậy \(D=\dfrac{1}{2012}\)
Nhớ tịk cho mink nhé
Ta có :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+..........+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+..........+\dfrac{1}{2013}\)
\(\Leftrightarrow S-P=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
\(1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2010}+\dfrac{1}{2012}\right)\)
\(\Rightarrow1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1005}+\dfrac{1}{1006}\right)\)
\(\Rightarrow\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow S=P\Rightarrow S-P=0\Rightarrow\left(S-P\right)^{2013}=1\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\left(\dfrac{2010}{2}+1\right)+\left(\dfrac{2009}{3}+1\right)+...+\left(\dfrac{1}{2011}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2011}+\dfrac{2012}{2012}}=\dfrac{1}{2012}\)
Bài 1:
Ta có:
\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)
\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)
Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:
\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)
Vậy......................
Chúc bạn học tốt!!!
Bài 2:
\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)
\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)
\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)
\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)
\(\Rightarrow-74< x< -16\)
\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)
Vậy..............................
Chúc bạn học tốt!!!
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+...+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)}\)
\(=\frac{1}{2014}\)
Vậy \(A=\frac{1}{2014}\)
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 ...
\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)
\(\Rightarrow\left(\dfrac{x+4}{2009}+1\right)+\left(\dfrac{x+3}{2010}+1\right)=\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+1}{2012}+1\right)\)
\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)
\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)
\(\Rightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)
=> x +2013 = 0
=> x = -2013
\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)
\(\Leftrightarrow\dfrac{x+4}{2009}+1+\dfrac{x+3}{2010}+1=\dfrac{x+2}{2011}+1+\dfrac{x+1}{2012}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)
\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2013=0\).Do \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
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 =))