e,
2x^2+y^2−6x+2xy−2y+5=0,2x^2+y^2−6x+2xy−2y+5=0
⇔(x^2−4x+4)+(x^2+2xy+y^2)−(2x+2y)+1=0⇔(x^2−4x+4)+(x^2+2xy+y2)−(2x+2y)+1=0
⇔(x−2)^2+(x+y)^2−2(x+y)+1=0⇔(x−2)^2+(x+y)^2−2(x+y)+1=0
⇔(x−2)^2+(x+y−1)^2=0⇔(x−2)^2+(x+y−1)^2=0
Mà (x−2)^2+(x+y−1)^2≥0∀x,y(x−2)^2+(x+y−1)^2≥0∀x,y
Suy ra xảy ra khi {(x−2)^2=0(x+y−1)^2=0{(x−2)^2=0(x+y−1)^2=0⇒{x−2=0,x+y−1=0⇒{x−2=0,x+y−1=0
⇒{x=2y+1=0⇒{x=2y+1=0⇒{x=2y=−1⇒{x=2y=−1
Vậy pt có nghiệm là (x;y)=(2;−1)