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a: \(\left(2x-3\right)^{2012}+\left(y-\dfrac{2}{5}\right)^{2014}+\left|x+y-z\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y-\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=\dfrac{2}{5}\\z=\dfrac{19}{10}\end{matrix}\right.\)
b: 2015-|x-2015|=x
=>|x-2015|=2015-x
=>x-2015<=0
hay x<=2015
d: |x-999|+|1998-2x|=0
=>x-999=0
hay x=999
Vì: \(\left(x-2014\right)^{10}\ge0,\forall x\)
\(\left(y-2015\right)^{100}\ge0,\forall y\)
\(\left(z-2016\right)^{1000}\ge0,\forall z\)
Mà: \(\left(x-2014\right)^{10}+\left(y-2015\right)^{100}+\left(z-2016\right)^{1000}=0\)
\(\Rightarrow\left\{\begin{matrix}x-2014=0\\y-2015=0\\z-2016=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=2014\\y=2015\\z=2016\end{matrix}\right.\)
vì ( x - 2014 )2014 \(\ge\)0 \(\forall\)x
( y - 2015 )2014 \(\ge\)0 \(\forall\)y
\(\Rightarrow\)( x - 2014 )2014 + ( y - 2015 )2014 \(\ge\)0 \(\forall\)x,y
Mà ( x - 2014 )2014 + ( y - 2015 )2014 = 0
\(\Rightarrow\)\(\hept{\begin{cases}\left(x-2014\right)^{2014}=0\\\left(y-2015\right)^{2014}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=2014\\y=2015\end{cases}}\)
Vậy ( x ; y ) = ( 2014 ; 2015 )
Vì (x-2014)2014 \(\ge\) 0
(y-2015)2014 \(\ge\)0
=> (x-2014)2014 + (y-2015)2014 \(\ge\) 0
Mà (x-2014)2014 + (y-2015)2014 = 0
=> \(\hept{\begin{cases}\left(x-2014\right)^{2014}=0\\\left(y-2015\right)^{2015}=0\end{cases}\Rightarrow\hept{\begin{cases}x-2014=0\\y-2015=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2014\\y=2015\end{cases}}}\)