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Ta thấy rằng: \(2^2>1\times2\) , \(3^2>2\times3\),..., \(2011^2>2010\times2011\).
\(\Rightarrow A< \frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{2011-2010}{2010\times2011}\)\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)\(=1-\frac{1}{2011}< 1.\)
Vậy A < 1.
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
S=\(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2010.2011.2012}\)
=\(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
=\(\frac{1}{2}-\frac{1}{2011.2012}< \frac{1}{2}\)(Vì \(\frac{1}{2011.2012}>0\))
=> S <\(\frac{1}{2}\)
\(S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{2010.2011.2012}\)
\(S=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2012-2010}{2010.2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2010.2011}-\frac{1}{2011.2012}\)
\(S=\frac{1}{1.2}-\frac{1}{2011.2012}=\frac{2023065}{4046132}\)
\(\text{Vì}\)\(\frac{2023065}{4046132}< \frac{1}{2}\Rightarrow S< P\)
\(\Rightarrow A=\frac{\left(2011^{2011}-3\right)+3}{2011^{2011}-3}=1+\frac{3}{2011^{2011}-3}\)
\(\Rightarrow B=\frac{2011^{2011}-1+3}{2011^{2011}-1}=1+\frac{3}{2011^{2011}-1}\)
Vì 20112011 - 3 < 20112011 - 1 =>\(\frac{3}{2011^{2011}-3}>\frac{3}{2011^{2011}-1}\)
\(\Rightarrow1+\frac{3}{2011^{2011}-3}>1+\frac{3}{2011^{2011}-1}\) hay A > B
