so sanh
A=2014 * 2014
B=2012 * 2016
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Ta có : \(\frac{2012}{2014}=\frac{1006}{1007}=1-\frac{1}{1007}\)
\(\frac{2014}{2016}=\frac{1007}{1008}=1-\frac{1}{1008}\)
Vì \(\frac{1}{1007}>\frac{1}{1008}\Rightarrow1-\frac{1}{1007}< 1-\frac{1}{1008}\Rightarrow\frac{2012}{2014}< \frac{2014}{2016}\)
a, A = 2012 . 2018
=> A = ( 2014 - 2 ) . 2018
=> A = 2014.2018 - 2.2018
b, B = 2014 . 2016
=> B = 2014 . ( 2018 - 2 )
=> B = 2014 . 2018 - 2014 .2
Vì 2.2018 > 2 .2014
=> A < B
\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)
\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)
\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)
\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)-xyz}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\)\(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)
\(\Leftrightarrow\)\(\left(xyz+y^2z\right)+\left(xyz+x^2z\right)+\left(xz^2+yz^2\right)+\left(xy^2+x^2y\right)=0\)
\(\Leftrightarrow yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(yz+xz+xy+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y\\x+z=0\end{cases}}=0\) hoặc y+z=0
Do đó ta có B=0
2014^2016 . 2015^2014 . 2013^2012 . 2012^2011
= (20142015 . 2014) . (20152014) .( 20132008 . 2013.2013.2013.203) .( 20122010 . 2012)
= (.....6)2015 . (...5) . (.........3)2008 . (...4)2010
= (....6).(....5).(.....6).(......4)
= .....0
ta có,A=(2012+2)*2014
=2014*2012+2014*2
B=2012*(2014+2)
=2012*2014+2012*2
Vì 2014*2<2012*2
Suy ra A>B