biết : \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}=0\)
vậy 4x+y-3z
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mình gợi ý nha
ta thấy biểu thức đầu \(\ge\)0
biểu thức 2\(\ge0\)
\(\Rightarrow\)biểu thức 3 =0
để vế trái =0
rồi lần lượt tìm xyz
Ta có (2x-1)\(^{2016}\)+(3y+6)\(^{2014}\)+(z-1)\(^{2012}\)=0
\(\Leftrightarrow\)(2x-1)\(^{2016}\)=0 ; (3y+6)\(^{2014}\)=0 ; (z-1)\(^{2012}\)=0
Ta co :(2x-1)\(^{2016}\)=0\(\Rightarrow\)2x-1=0\(\Rightarrow\)2x=1\(\Rightarrow\)x=\(\frac{1}{2}\)
(3y+6)\(^{2014}\)=0 \(\Rightarrow\)3y+6=0 \(\Rightarrow\)3y=-6 \(\Rightarrow\)y=-2
(z-1)\(^{2012}\)=0 \(\Rightarrow\)z-1=0 \(\Rightarrow\)z=1
Vậy 4x+y-3z=4*\(\frac{1}{2}\)+(-2)-3*1=2-2-3=-3
Ta có: \(x+2y+3x=0\Leftrightarrow x=-\left(2y+3z\right)\)
Lại có: \(2xy+6yz+3xz=0\Leftrightarrow x\left(2y+3z\right)+6yz=0\)
\(\Leftrightarrow-\left(2y+3z\right)\left(2y+3z\right)+6yz=0\Leftrightarrow-\left(2y+3z\right)^2+6yz=0\)
\(\Leftrightarrow\left(2y+3z\right)^2-6yz=0\Leftrightarrow4y^2+12yz+9z^2-6yz=0\)
\(\Leftrightarrow4y^2+6yz+9z^2=0\Leftrightarrow\left(2y+\dfrac{3z}{2}\right)^2+\dfrac{27z^2}{4}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2y+\dfrac{3z}{2}\right)^2=0\\\dfrac{27z^2}{4}=0\end{matrix}\right.\) \(\Rightarrow y=z=0\Rightarrow x=0\)
\(\Rightarrow S=\dfrac{\left(-1\right)^{2019}-1^{2017}+\left(-1\right)^{2015}}{1^{2018}+2.0^{2016}+0^{2014}+2}=\dfrac{-1-1+-1}{1+0+0+2}=\dfrac{-3}{3}=-1\)
Vì \(\left(2x-1\right)^{2016}\ge0;\left(3y+6\right)^{2014}\ge0;\left(z-1\right)^{2012}\ge0\)
\(\Rightarrow\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}\ge0\)
Để \(\left(2x-1\right)^{2016}+\left(3y+6\right)^{2014}+\left(z-1\right)^{2012}=0\)\(\Leftrightarrow\left(2x-1\right)^{2016}=0;\left(3y+6\right)^{2014}=0;\left(z-1\right)^{2012}=0\)
\(\Leftrightarrow2x-1=0;3y+6=0;z-1=0\)
\(\Rightarrow x=\dfrac{1}{2};y=-2;z=1\)
\(\Rightarrow4x+y-3z=4.\dfrac{1}{2}+\left(-2\right)-3.1=2-2-3=-3\)
giúp mk vs nha mina !!!