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$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
a)\(\frac{2013}{2015}< \frac{2014}{2016}\)
b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
Ta có:
\(B=\frac{2015+2016}{2016+2017}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
vì: \(\frac{2015}{2016}>\frac{2015}{2016+2017}\)VÀ \(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Vậy: \(A>B\)
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
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\(A=\frac{2010}{2009}+\frac{2011}{2010}+\frac{2012}{2011}+\frac{2009}{2012}=\left(1+\frac{1}{2009}\right)+\left(1+\frac{1}{2010}\right)+\left(1+\frac{1}{2011}\right)+\frac{2009}{2012}>\left(1+\frac{1}{2012}\right)+\left(1+\frac{1}{2012}\right)+\left(1+\frac{1}{2012}\right)+\frac{2009}{2012}=\left(1+1+1\right)+\left(\frac{1}{2012}+\frac{1}{2012}+\frac{1}{2012}+\frac{2009}{2012}\right)=3+1=4\)Vì 1/2009,1/2010,1/2011>1/2012
Vậy A>4