x^3+y^3+9xy=0
<=>x^3+3.x^2.y+3.x.y^2+y^3-(3.x^2.y+3.x.y^2+9xy)=0
<=>(x+y)^3-3xy(x+y+3)=0
<=>(x+y)^3+27-3xy(x+y+3)=27
<=>(x+y+3)[(x+y)^2+3(x+y)+9]-3xy(x+y+3)=27
<=>(x+y+3)(x^2-xy+y^2+3x+3y+9)=27=3.9=9.3=27.1=1.27
ta có: x,y>=0
=>x+y+3>=3 (3)
=>3x+3y+9>=9 (1)
ta lại có: x^2-xy+y^2...