(1) ⇔(x+y)2−2xyxy+2x+y−1xy=0 ⇔(x+y)2−1xy+2x+y−2=0 ⇔(x+y−1)(x+y+1)xy+2(1−x−y)xy=0 ⇔x+y=1 or x2+y2+x+y=0 +) x+y=1⇒3x2−4x−1=0⇔x=2±√73+) x2+y2=−(x+y)⇒1+2x−x2=x2+y2+1x2+y2≥2⇔x=1⇒(x;y)=.....
(1)
⇔(x+y)2−2xyxy+2x+y−1xy=0 ⇔(x+y)2−1xy+2x+y−2=0 $\Leftrightarrow \frac{(x+y-1)(x+y+1)}{xy} +\frac{2(1-x-y)}{x
+y}=0
\Leftrightarrow x+y=1
orx^{2}+y^{2}+x+y=0
+)x+y=1
⇒3x2−4x−1=0⇔x=2±√73$+)$x2+y2=−(x+y)\Rightarrow 1+2x-x^{2}=x^{2}+y^{2}+\frac{1}{x^{2}+y^{2}}\geq2
⇔x=1\Rightarrow (x;y)=.....$