√(3x+4) + y^2-6y+9=0
√(3x+4) + (y-3)^2=0
√(3x+4)≥0,(y-3)^2≥0
√(3x+4)=0,(y-3)^2=0
x=-4/3,y=3
-xy = -(-4/3)*3 = 4
-xy开方= ±√4 = ±2谢谢了