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Ta có:
Q=2010/2011+2012+2013+2011/2011+2012+2013+2012/2011+2012+2013
Mà 2010/2011+2012+2013<2010/2011
2011/2011+2012+2013<2011/2012
2012/2011+2012+2013<2012/2013
=>Q<P
cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)
\(=x-1=2013-1=2012\)
Có : \(\frac{2011}{2012}=\frac{2012-1}{2012}=1-\frac{1}{2012}\)
Có : \(\frac{2012}{2013}=\frac{2013-1}{2013}=1-\frac{1}{2013}\)
Có : \(\frac{2013}{2011}=\frac{2011+2}{2011}=1+\frac{2}{2011}\)
Cộng vế với vế ta có : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}=1-\frac{1}{2012}+1-\frac{1}{2013}+1+\frac{2}{2011}=1+1+1-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)=3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}>0\) nên \(3-\left(\frac{1}{2012}+\frac{1}{2013}-\frac{2}{2011}\right)<3\)
Vậy \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}<3\)
A=\(\frac{2014}{2014^a}+\frac{2014}{2014^b}\)=B=\(\frac{2013}{2015^a}\)+\(\frac{2015}{2013^b}\)
Ta có: 2014/\(2014^a\)+2014/2014^b= 2013/2014^a + 1/2014^a +2015/2014^a - 1/2014^a
=(2013/2014^a + 2015/2014^b) + ( 1/2014^a + 1/2014^b)
= B + (1/2014^a + 1/2014^b)
*Nếu a=b thì A=B
*Nếu a>b thì (1/2014^a + 1/2014^b) >0
\(\Rightarrow\) A< B
*Nếu a<b thì (1/2014^a + 1/2014^b)>0
\(\Rightarrow\) A>B
P=2013-2013=0