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A = \(\frac{2012-1}{\sqrt{2012}}+\frac{2011+1}{\sqrt{2011}}=\sqrt{2012}-\frac{1}{\sqrt{2012}}+\sqrt{2011}+\frac{1}{\sqrt{2011}}\)
A = \(\sqrt{2012}+\sqrt{2011}+\left(\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)=B+\left(\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
Mà 2011 < 2012 nên \(\frac{1}{\sqrt{2011}}>\frac{1}{\sqrt{2012}}\Rightarrow\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}>0\)
=> A > B
Giải:
(1+1/2!)+(1+2/3!)+(1+3/4!)+....+(1+2011/2012!)=2011+(1/2!+2/3!+3/4!+...+2011/2012!)
=2011+(\(\frac{1}{2!}\)+\(\frac{3-1}{3!}\)+\(\frac{4-1}{4!}\)+...+\(\frac{2012-1}{2012!}\))= 2011 +(\(\frac{1}{2!}\)+\(\frac{1}{2!}\)-\(\frac{1}{3!}\)+\(\frac{1}{3!}\)-\(\frac{1}{4!}\)+...+\(\frac{1}{2011!}\)-\(\frac{1}{2012!}\))
= 2011+(1-\(\frac{1}{2012!}\))=2012 - \(\frac{1}{2012!}\)<2012 (đpcm)
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}\)
\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2012}}+\dfrac{1}{3^{2013}}\)
\(\Rightarrow A-\dfrac{1}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{2012}}-\dfrac{1}{3^{2013}}\)\(\Rightarrow\dfrac{2}{3}A=\dfrac{1}{3}-\dfrac{1}{3^{2013}}< \dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}A< \dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{1}{3}.\dfrac{3}{2}=\dfrac{1}{2}\)
Vậy \(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2011}}+\dfrac{1}{3^{2012}}< \dfrac{1}{2}\)
Đặt \(\sqrt{2011}=a;\sqrt{2012}=b\)
Theo đề, ta có: \(A=\dfrac{a^2}{b}+\dfrac{b^2}{a}=\dfrac{a^3+b^3}{ab}\)
B=a+b
\(A-B=\dfrac{a^3+b^3}{ab}-\left(a+b\right)=\dfrac{a^3+b^3-a^2b-ab^2}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a-b\right)^2}{ab}>0\)
=>A>B