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24 tháng 9 2016
Mẫu số của A \(=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
\(=\left(1+1+...+1\right)+\left(\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\right)\)
(2012 số 1) (2011 phân số)
\(=\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(=2013.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
=> \(A=\frac{1}{2013}\)
24 tháng 9 2016
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}\)
\(\Rightarrow A=\frac{1}{2013}\)
Vậy \(A=\frac{1}{2013}\)
6 tháng 2 2017
\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\left(\frac{2012}{2}+1\right)+\left(\frac{2011}{3}+1\right)+....+\left(\frac{1}{2013}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}}{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2003}\right)}\)
\(=\frac{1}{2014}\)
S
6 tháng 7 2018
a, ĐK: \(x+1\ge0\Leftrightarrow x\ge-1\)
Ta có: |3-2x|=x+1
=>\(\orbr{\begin{cases}3-2x=x+1\\3-2x=-x-1\end{cases}\Rightarrow\orbr{\begin{cases}x+2x=3-1\\-x+2x=3+1\end{cases}\Rightarrow}\orbr{\begin{cases}3x=2\\x=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\left(tmđk\right)\\x=4\left(tmđk\right)\end{cases}}}\)
Vậy x=2/3 hoặc x=4
b, Xét VP ta có: \(\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(=1+\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}+1\)
\(=\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2013}+\frac{2014}{2014}=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)\)
=>\(x=\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}=2014\)
Vậy x=2014
1 tháng 3 2016
TRƯỚC TIÊN BẠN NÊN XÉT MẪU SỐ :
*Xét mẫu số :
\(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
\(=\left(1+1+1+...+1\right)+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
(2012 SỐ HẠNG 1) (2012 SỐ HẠNG LÀ PHÂN SỐ)
\(=\left(1+\frac{2012}{2}\right)+\left(1+\frac{2011}{3}\right)+\left(1+\frac{2012}{4}\right)+...+\left(1+\frac{1}{2013}\right)\)
(2012 SỐ HẠNG (NHÓM))
\(=\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+.....+\frac{2014}{2013}=2014\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}=\frac{1}{2014}\)
22 tháng 6 2017
Ta có:
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}=\frac{1}{1\cdot2:2}+\frac{1}{2\cdot3:2}+...+\frac{1}{2012\cdot2013:2}\)
\(=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{2012\cdot2013}=2\left[\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2012\cdot2013}\right]\)
\(=2\left[\left[\frac{1}{1}-\frac{1}{2}\right]+\left[\frac{1}{2}-\frac{1}{3}\right]+\left[\frac{1}{3}-\frac{1}{4}\right]+...+\left[\frac{1}{2012}-\frac{1}{2013}\right]\right]\)
\(=2\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right]=2\left[1-\frac{1}{2013}\right]\)
\(=2\cdot\frac{2012}{2013}=\frac{4024}{2013}\)
Thế vào bài toán, ta có:
\(\frac{2\cdot2012}{\frac{4024}{2013}}=\frac{4024}{\frac{4024}{2013}}=2013\)
\(A=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}}\)
\(A=\frac{4024}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2012.2013:2}}\)
\(A=\frac{4024}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{2013}\right)}\)
\(A=\frac{4024}{1+1-\frac{2}{2013}}=\frac{4024}{2-\frac{2}{2013}}=4024:\frac{4024}{2013}=\frac{4024.2013}{4024}=2013\)