1. x2 + 2.\(\dfrac{1}{2}\)xy + \(\dfrac{1}{4}\)y2 + \(\dfrac{3}{4}\)y2 = 0
(=) (x+\(\dfrac{1}{2}\)y)2 +\(\dfrac{3}{4}\)y2 = 0
(=) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}y\\y=0\end{matrix}\right.\) (=) x=y=0
2.
(x2 + 2xy +y2) + (x2 -2x +1) = 0
(=) (x+y)2 + (x-1)2 = 0
(=)\(\left\{{}\begin{matrix}x=-y\\x=1\end{matrix}\right.\) (=) \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
1 ) \(x^2+xy+y^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)=0\)
\(\Leftrightarrow x^3-y^3=0\)
\(\Leftrightarrow x^3=y^3\)
\(\Leftrightarrow x=y\)
\(\Leftrightarrow x^2+xy+y^2=x^2+x^2+x^2\)
\(\Leftrightarrow0=3x^2\)
\(\Leftrightarrow x^2=0\)
\(\Leftrightarrow x=0\)
\(\Leftrightarrow y=0\)
Vậy \(x=y=0\)
2 ) \(2x^2+y^2+2xy-2x+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x-1\right)^2=0\left(1\right)\)
Do \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\forall x;y\\\left(x-1\right)^2\ge0\forall x\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2\ge0\forall x;y\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)
Vậy \(x=1;y=-1\)