Rút gọn biểu thức sau:\(4-\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{3+\frac{1}{2+\frac{1}{4+\frac{1}{2}}}}}}}\)
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8 tháng 4 2016
Xét mẫu số ta có: \(2012+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}\)
=\(2012+\left(\frac{2014-2}{2}+\frac{2014-3}{3}+...+\frac{2014-2013}{2013}\right)\)
= \(2012+\left(\frac{2014}{2}+\frac{2014}{3}+\frac{2014}{4}+...+\frac{2014}{2013}\right)-\left(\frac{2}{2}+\frac{3}{3}+\frac{4}{4}+...+\frac{2013}{2013}\right)\)
= \(2012+2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2012\)
= \(2014\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)\)
\(\Rightarrow A=\frac{1}{2014}\)
2 tháng 8 2016
CM : \(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)^2}=1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
= \(\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}=\frac{n^2\left[\left(n+1\right)^2+1\right]+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{n^2\left(n^2+2n+2\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
=\(\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\) = \(\frac{\left(n^2+n+1\right)^2}{\left(n^2+n\right)^2}\) =>\(\sqrt{\left(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\right)}=\frac{n^2+n+1}{n^2+n}\)
\(=1+\frac{1}{n^2+n}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có :
A = \(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+\left(1+\frac{1}{4}-\frac{1}{5}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)
= 2012 - \(\frac{1}{2013}\) \(\approx\) 2012
SL
30 tháng 3 2016
ở mẫu n4+n2+1=(n2+n+1)(n2-n+1)
\(\frac{2n}{n^4+n^2+1}=\frac{\left(n^2+n+1\right)-\left(n^2-2+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)
