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Ta có \(\sqrt{a^{2012}+2011}\le\dfrac{a^{2012}+2011+1}{2}\)
\(\Leftrightarrow\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}\ge\dfrac{a^{2012}+2012}{\dfrac{a^{2012}+2012}{2}}=2\)
Dấu \("="\Leftrightarrow a^{2012}+2011=1\Leftrightarrow a\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
\(\Rightarrow\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}>2\)
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\)
\(< \sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(\Rightarrow N< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}\right)\)
\(N< 2\left(1-\frac{1}{\sqrt{2012}}\right)< 2.1=2\)
\(\sqrt{\left(1-\sqrt{2012}\right)^2}\sqrt{2013+2\sqrt{2012}}\)
\(=\sqrt{\left(1-2\sqrt{503}\right)^2}\sqrt{\left(1+\sqrt{2012}\right)^2}\)
\(=\left(2\sqrt{503}-1\right)\left(1+\sqrt{2012}\right)\)
\(=\left(2\sqrt{503}-1\right)\left(1+2\sqrt{503}\right)\)
\(=\left(2\sqrt{503}-1\right)\left(2\sqrt{503}+1\right)\)
\(=4\cdot503-1\)
\(=2012-1\)
\(=2011\)
\(A=\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|\)
\(=\left|x-2012\right|+\left|2014-x\right|+\left|x-2010\right|\)
\(\ge\left|x-2012\right|+\left|2014-x+x-2010\right|\)
\(=\left|x-2012\right|+4\)
Vì \(\left|x-2012\right|\ge0\forall x\)
\(\Rightarrow\left|x-2012\right|+4\ge4\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=2012\)
Vậy MIN \(A=4\Leftrightarrow x=2012\)
