2012+2013x2014
2014x2015+2016
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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)-xyz}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\)\(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)
\(\Leftrightarrow\)\(\left(xyz+y^2z\right)+\left(xyz+x^2z\right)+\left(xz^2+yz^2\right)+\left(xy^2+x^2y\right)=0\)
\(\Leftrightarrow yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(yz+xz+xy+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y\\x+z=0\end{cases}}=0\) hoặc y+z=0
Do đó ta có B=0
1) (1998.1999 - 1998.1998 - 1000 - 98) : 2016
= [1998.(1999 - 1998) - 1098) : 2016
= (1998.1 - 1098) : 2016
= (1998 - 1098) : 2016
= 900 : 2016
= \(\frac{25}{56}\)
b) 2016.2016 - 2012.2020
= 20162 - (2016 - 4).(2016 + 4)
= 20162 - (20162 - 42)
= 20162 - 20162 + 42
= 0 + 16
= 16
\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)
\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)
\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)
\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)
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