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\(x^2+y^2+z^2+2xy+2yz+2xz+x^2-2x+1+y^2+2y+1=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x-1=0\\y+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=-\left(x+y\right)\\x=1\\y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=0\end{matrix}\right.\)
\(\Rightarrow A=1^{2018}+\left(-1\right)^{2018}+0^{2018}=1+1+0=2\)
2x2 + 2y2 + 3xy - x + y + 1 = 0
2x2 + 2y2 + 4xy - xy - x + y + 1 = 0
(2x2 + 2y2 + 4xy) + (-xy - x) + (y + 1) = 0
2(x + y)2 - x(y + 1) + (y + 1) = 0
2(x + y)2 + (y + 1)(1 - x) = 0
Do (x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0
\(\Rightarrow y+1=0;1-x=0\)
*) y + 1 = 0
y = -1
*) 1 - x = 0
x = 1
Với x = 1; y = -1, ta có:
B = [1 + (-1)]2018 + (1 - 2)2018 + (-1 - 1)2018
= 1 + 22018
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
Theo đề bài : 2x2 + 2y2 + 2xy - 2x + 2y + 2 = 0
\(\Rightarrow\) ( x2 + 2xy + y2 ) + ( x2 - 2x + 1 ) + ( y2 + 2y + 1 ) = 0
( x + y )2 + ( x - 1 )2 + ( y + 1 )2 = 0
Ta thấy : \(\left(x+y\right)^2\ge0;\forall x,y\in R\)
\(\left(x-1\right)\ge0;\forall x\in R\)
\(\left(y+1\right)^2\ge0;\forall y\in R\)
\(\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0;\forall x,y\in R\)
Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x-1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\left(\text{Thỏa mãn}\right)\)
Thay \(x=1\) và \(y=-1\) vào \(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\) , ta được :
\(A=\left(x-2\right)^{2017}+\left(y+1\right)^{2018}\)
\(A=\left(1-2\right)^{2017}+\left(-1+1\right)^{2018}\)
\(A=-1+0\)
\(A=-1\)
Vậy \(A=-1\Leftrightarrow\left\{{}\begin{matrix}2x^2+2y^2+2xy-2x+2y+2=0\\x=1\\y=-1\end{matrix}\right.\)
\(2x^2+2y^2+z^2-2x+2y+2xy+2yz+2zx+2=0\)
\(\Leftrightarrow\)\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\)\(\left(x+y\right)^2+\left(y+z\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\)\(x=-y=z=1\)
\(\Rightarrow\)\(A=x^{2018}+y^{2018}+z^{2018}=1^{2018}+\left(-1\right)^{2018}+1^{2018}=3\)
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