Ta có : \(\left(11x+6y+2015\right)\left(x-y+3\right)=0\)
Mà \(x,y>0\)
=> \(11x+6y+2015>0\)
=> \(x-y+3=0\)
=> \(y=x+3\)
Ta có : \(P=x\left(x+3\right)-5x+2016\)
=> \(P=x^2+3x-5x+2016\)
=> \(P=x^2-2x+2015=\left(x-1\right)^2+2015\)
Ta thấy : \(\left(x-1\right)^2\ge0\forall x\)
=> \(\left(x-1\right)^2+2015=P\ge2015\forall x\)
Vậy MinP = 2015 <=> x = 1 ( y = 4 )