19. [10] Define $x \star y=\frac{\sqrt{x^{2}+3 x y+y^{2}-2 x-2 y+4}}{x y+4}$. Compute
$$
((\cdots((2007 \star 2006) \star 2005) \star \cdots) \star 1) .
$$
Answer: $\frac{\sqrt{15}}{9}$. Note that $x \star 2=\frac{\sqrt{x^{2}+6 x+4-2 x-4+4}}{2 x+4}=\frac{\sqrt{(x+2)^{2}}}{2(x+2)}=\frac{1}{2}$ for $x>-2$. Because $x \star y>0$ if $x, y>0$, we need only compute $\frac{1}{2} \star 1=\frac{\sqrt{\frac{1}{4}+\frac{3}{2}+1-3+4}}{\frac{1}{2}+4}=\frac{\sqrt{15}}{9}$.