CMR:1-1/2^2-1/3^2-......-1/2011^2>1/2011
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Từ giả thiết ta suy ra \(\frac{1}{a_1}-1=\frac{a_2+\cdots+a_{2011}}{a_1}\ge\frac{2010\sqrt[2010]{a_2\cdots a_{2011}}}{a_1}=\frac{2010\left(\sqrt[2010]{\frac{a_1\cdots a_{2011}}{a_1}}\right)}{a_1}.\)
Tương tự, ta thiết lập 2010 bất đẳng thức còn lại cho \(\frac{1}{a_2}-1,\ldots,\frac{1}{a_{2011}}-1\) rồi nhân vào ta sẽ thu được
\(\left(\frac{1}{a_1}-1\right)\left(\frac{1}{a_2}-1\right)\cdots\left(\frac{1}{a_{2012}}-1\right)\ge\frac{2010^{2011}\left(\sqrt[2010]{\frac{a_1\cdots a_{2011}}{a_1}}\right)\cdots\left(\sqrt[2010]{\frac{a_1\cdots a_{2011}}{a_{2011}}}\right)}{a_1\cdots a_{2011}}=2010^{2011}\)
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Ta có :\(\left(2011+1\right)^2=2011^2+1+2.2011\)
\(\Rightarrow2011^2+1=2012-2.2011\)
\(\Rightarrow N=\sqrt{2012^2-2.2011+\left(\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=\sqrt{\left(2012-\dfrac{2011}{2012}\right)^2}+\dfrac{2011}{2012}\)
\(=2012-\dfrac{2011}{2012}+\dfrac{2011}{2012}\)
\(=2019\)
Vậy N có giá trị là một số tự nhiên.
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Ta có: \(2.S=2.\left(\frac{1}{1^4+1^2+1}+...+\frac{2011}{2011^4+2011^2+1}\right)\)
Xét hạng tử tống quát: \(\frac{2.n}{n^4+n^2+1}=\frac{2.n}{\left(n^4+2n^2+1\right)-n^2}=\frac{\left(n^2+n+1\right)-\left(n^2-n+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)\(=\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\)
Từ đó: \(\frac{2.1}{1^4+1^2+1}=\frac{1}{1}-\frac{1}{3}\)
\(\frac{2.2}{2^4+2^2+1}=\frac{1}{3}-\frac{1}{7}\)
.....
\(\frac{2.2011}{2011^4+2011^2+1}=\frac{1}{4042111}-\frac{1}{4046133}\)
Từ đó => 2.S= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+...+\frac{1}{4042111}-\frac{1}{4046133}\)=\(1-\frac{1}{4046133}\)=\(\frac{4046132}{4046133}\)
=> S\(=\frac{2023066}{4046133}\)
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\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left[\frac{1}{x^2}-\frac{2}{xy}+\frac{1}{y^2}\right]-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2}{xy}:\left[\frac{y^2-2.xy+x^2}{x^2y^2}\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}.\frac{x^2y^2}{x^2-2xy+y^2}-\frac{x^2+y^2}{x^2-2xy+y^2}\)
\(=\frac{2xy}{x^2-2xy+y^2}+\frac{-x^2-y^2}{x^2-2xy-y^2}\)
\(=\frac{2xy-x^2-y^2}{x^2-2xy+y^2}=\frac{-\left(x^2-2xy+y^2\right)}{x^2-2xy+y^2}=-1\)
\(\frac{2011^3+11^3}{2011^3+2000^3}\)
\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)
\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)
\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)
\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)
đpcm
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Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)
B=2012.A
=>A/B=1/2012
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\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow S< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
\(\Rightarrow S< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)
S = \(1-\frac{1}{2^2}-\frac{1}{3^2}-....-\frac{1}{2011^2}<1-\frac{1}{2.3}-\frac{1}{3.4}-.....-\frac{1}{2011.2012}\)
Đặt A = \(-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{2011.2012}=-\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2011.2012}\right)\)
\(A=-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{2012}\right)=-\left(\frac{1}{2}-\frac{1}{2012}\right)=-\frac{1005}{2012}\)
S = 1 + \(\left(-\frac{1005}{2012}\right)=\frac{1007}{2012}>\frac{1}{2011}\)
=> ĐPCM