Xét $4x^2+4z^2-2.4y(x+2)+20y^2+16z=17-10-27=-20$
$⇒4x^2+4z^2-8xy-16y+20y^2+16z+20=0$
$⇒(4x^2-8xy+4y^2)+(4z^2+16z+16)+(16y^2-16y+4)=0$
$⇒4.(x-y)^2+4.(z+2)^2+4.(2y-1)^2=0$
Ta có: $(x-y)^2+(z+2)^2+(2y-1)^2≥0∀x;y;z$
Dấu $=$ xảy ra $⇔x-y=0;z+2=0;2y-1=0⇔x=y=\dfrac{1}{2};z=-2$
Khi đó: $M=10.\dfrac{1}{2}+4.\dfrac{1}{2}+2019.(-2)=-4031$