Tính \(\frac{2012.2013-1}{2012^2+2011}\)
K
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3 tháng 9 2015
\(A=\left(1-\frac{1}{2011}\right)-\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)-\left(1-\frac{1}{2014}\right)\)
\(=1-\frac{1}{2011}-1+\frac{1}{2012}+1-\frac{1}{2013}-1+\frac{1}{2014}\)
\(=\left(1-1+1-1\right)-\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}\right)\)
còn lại bó tay @@
3 tháng 9 2015
\(A=\frac{2010}{2011}-\frac{2011}{2012}+\frac{2012}{2013}-\frac{2013}{2014}\)
và
\(B=\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
TA
27 tháng 7 2016
Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
\(=\frac{\sqrt{n+1}}{\sqrt{n}.\sqrt{n+1}}-\frac{\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thay n = 1, 2, 3, ..., 2011 vào C ta có:
\(C=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
Vậy \(C=1-\frac{1}{\sqrt{2012}}.\)
27 tháng 3 2018
\(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
\(P=\left(1+\frac{1}{3}+...+\frac{1}{2011}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(P=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}-1-\frac{1}{2}-...-\frac{1}{1006}\)
\(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\) (1)
\(Q=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\) (2)
\(\left(1\right)\left(2\right)\Rightarrow\frac{P}{Q}=\frac{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}=1\)
G
4 tháng 3 2018
Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
bài này dễ mà bn
ta Có \(\frac{2012.2013-1}{2012^2+2011}=\)\(\frac{2012.\left(2012+1\right)-1}{2012^2+2011}=\)\(\frac{2012^2+2012-1}{2012^2+2011}=\)\(\frac{2012^2+2011}{2012^2+2011}=1\)