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A=2011^2012-2011^2011= 2011^2011 * 2011 -2011^2011= 2011^2011 *(2011-1)= 2011^2011 *2010
B=2011^2013-2011^2012=2011^2012*2011- 2011^2012= 2011^2012 *(2011-1) = 2011^2012 *2010
vì 2011^2011*2010 < 2011^2012*2010 nên A<B
Ta có : 2011^2013 x M = (2010^2012 x 2011 + 2011^2013)^2013 > (2010^2013 + 2011^2013)^2013 = N x (2010^2013 + 2011^2013)
Do đó: 2011^2013 x M > N x (2010^2013 + 2011^2013)
<=> M > N x [(2010/2011)^2013 + 1] ==> M > N (điều phải chứng minh)
Ta có: \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)\(\Rightarrow A>B\)
So sánh: \(\frac{2010}{2011}+\frac{2011}{2012}\) với \(\frac{2010+2011}{2011+2012}\)
\(P=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}=\frac{2010}{8144863716}+\frac{2011}{8144863716}+\frac{2012}{8144863716}\)
\(=\frac{6033}{8144863716}=\frac{1}{1350052}\)
\(Q=2010+2011+\frac{2012}{2011}+2012+2013\)
\(=2010+2011+2012+2013+\frac{2012}{2011}\)
\(=8046+\frac{2012}{2011}=\frac{8046}{1}+\frac{2012}{2011}\)
\(=\frac{16180506}{2011}+\frac{2012}{2011}=\frac{16182518}{2011}\)
\(\frac{2010+2011+2012}{2011+2012+2013}=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Vì \(\frac{2010}{2011+2012+2013}<\frac{2010}{2011};\frac{2011}{2011+2012+2013}<\frac{2011}{2012};\frac{2012}{2011+2012+2013}<\frac{2012}{2013}\)
nên phép dưới nhỏ hơn phép trên
Ta có A=2010/2011+2011/2012+2012/2010
= (2010/2011+1/2011)+1+(2011/2012+1/2012)
=1+1+1=3
=> A=3
tích tui ik