Tính giá trị biểu thức :
\(A=\frac{\frac{1}{2013}+\frac{2}{2012}+\frac{3}{2011}+...+\frac{2011}{3}+\frac{2012}{2}+\frac{2013}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}}\)
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26 tháng 1 2016
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HT
18 tháng 9 2015
\(S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}}\)
Xét mẫu:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...+\frac{1}{2014}\)
= \(\left(1+\frac{2013}{2}\right)+\left(1+\frac{2012}{3}\right)+...+\left(1+\frac{1}{2014}\right)+1\)
= \(\frac{2014}{2}+\frac{2014}{3}+....+\frac{2014}{2013}+\frac{2014}{2014}\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}}{2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)}\)
\(\Rightarrow S=\frac{1}{2014}\)
HP
24 tháng 10 2020
\(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
Ta có \(f\left(x\right)+f\left(1-x\right)=1\) khi đó
\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+...+\left[f\left(\frac{1005}{2012}\right)+f\left(\frac{1007}{2012}\right)\right]+f\left(\frac{1006}{2012}\right)\)
\(=1+1+...+1+f\left(\frac{1}{2}\right)=1005+\frac{\left(\frac{1}{2}\right)^3}{1-3.\frac{1}{2}+3.\left(\frac{1}{2}\right)^2}=1005+\frac{1}{2}=\frac{2011}{2}\)
24 tháng 10 2020
Ta có: \(F\left(x\right)=\frac{x^3}{1-3x+3x^2}\)
\(\Leftrightarrow F\left(1-x\right)=1-\frac{x^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)
\(=\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
Ta có: \(F\left(x\right)+F\left(1-x\right)\)
\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3x+3x^2}\)
\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)
\(\Leftrightarrow F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)=1\)
...
\(F\left(\frac{1005}{2012}\right)+F\left(\frac{1007}{2012}\right)=1\)
Do đó: \(A=F\left(\frac{1}{2012}\right)+F\left(\frac{2}{2012}\right)+...+F\left(\frac{2010}{2012}\right)+F\left(\frac{2011}{2012}\right)\)
\(=\left[F\left(\frac{1}{2012}\right)+F\left(\frac{2011}{2012}\right)\right]+\left[F\left(\frac{2}{2012}\right)+F\left(\frac{2010}{2012}\right)\right]+...+F\left(\frac{1006}{2012}\right)\)
\(=1+1+...+F\left(\frac{1}{2}\right)\)
\(=1005+\left[\left(\frac{1}{2}\right)^3:\left(1-3\cdot\frac{1}{2}+3\cdot\frac{1}{4}\right)\right]\)
\(=1005+\left[\frac{1}{8}:\left(1-\frac{3}{2}+\frac{3}{4}\right)\right]\)
\(=1005+\left(\frac{1}{8}:\frac{1}{4}\right)\)
\(=1005+\frac{1}{2}=\frac{2011}{2}\)
4 tháng 3 2018
A=\(\frac{1+\frac{2011}{2}+1+\frac{2010}{3}+1+...+\frac{1}{2012}+1+1}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=\(\frac{2013\left(\frac{1}{2}+...+\frac{1}{2013}\right)}{\frac{1}{2}+...+\frac{1}{2013}}\)
A=2013
Mà 2013: 3 = 671
Vậy A : 3 dư 0 hay\(A⋮3\)
Ta có: Tử là:
B=\(\frac{1}{2013}+\frac{2}{2012}+...+\frac{2012}{2}+\left(1+1+...+1\right)\) (2013 số hạng 1)
=\(\left(\frac{1}{2013}+1\right)+\left(\frac{2}{2012}+1\right)+...+\left(\frac{2012}{2}+1\right)+\left(1\right)\)
=\(\frac{2014}{2013}+\frac{2014}{2012}+...+\frac{2014}{2}+\frac{2014}{2014}\)
=\(2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=>A=\(\frac{2014\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2014
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