2011/2010 *2012/2011 * 2013/2012 * 2014/2013 * 1005/1007
dấu * là dấu nhân
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$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$
$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$
Ta thấy
\(\dfrac{2010}{2011}< 1\)
\(\dfrac{2011}{2012}< 1\)
\(\dfrac{2012}{2013}< 1\)
\(\dfrac{2013}{2014}< 1\)
=> Tổng M của những phân số trên sẽ nhỏ hơn 1
=> M < 1
\(A=\left(1-\frac{1}{2011}\right)-\left(1-\frac{1}{2012}\right)+\left(1-\frac{1}{2013}\right)-\left(1-\frac{1}{2014}\right)\)
\(=1-\frac{1}{2011}-1+\frac{1}{2012}+1-\frac{1}{2013}-1+\frac{1}{2014}\)
\(=\left(1-1+1-1\right)-\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2014}\right)\)
còn lại bó tay @@
\(A=\frac{2010}{2011}-\frac{2011}{2012}+\frac{2012}{2013}-\frac{2013}{2014}\)
và
\(B=\frac{1}{2010.2011}-\frac{1}{2012.2013}\)
\(P=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}=\frac{2010}{8144863716}+\frac{2011}{8144863716}+\frac{2012}{8144863716}\)
\(=\frac{6033}{8144863716}=\frac{1}{1350052}\)
\(Q=2010+2011+\frac{2012}{2011}+2012+2013\)
\(=2010+2011+2012+2013+\frac{2012}{2011}\)
\(=8046+\frac{2012}{2011}=\frac{8046}{1}+\frac{2012}{2011}\)
\(=\frac{16180506}{2011}+\frac{2012}{2011}=\frac{16182518}{2011}\)
\(\frac{2011}{2010}\times\frac{2012}{2011}\times\frac{2013}{2012}\times\frac{2014}{2013}\times\frac{1005}{1007}\)
\(=\frac{2014}{2010}\times\frac{1005}{1007}\)
\(=\frac{2\times1007\times1005}{2\times1005\times1007}\)
\(=1\)
\(\frac{2011}{2010}\cdot\frac{2012}{2011}\cdot\frac{2013}{2012}\cdot\frac{2014}{2013}\cdot\frac{2010}{2014}\)
\(=\frac{2010\cdot2011\cdot2012\cdot2013\cdot2014}{2010\cdot2011\cdot2012\cdot2013\cdot2014}\)
= 1