8．已知正数$x,y$满足$\frac{8}{3x^{2} + 2xy} + \frac{3}{xy + 2y^{2}} = 1$，则$xy$的最小值是\\\\\\\\\_．
【详解】根据题意，由$\frac{8}{3x^{2} + 2xy} + \frac{3}{xy + 2y^{2}} = 1$可得$\frac{8(xy + 2y^{2}) + 3(3x^{2} + 2xy)}{(3x^{2} + 2xy)(xy + 2y^{2})} = 1$，
即$16y^{2} + 9x^{2} + 14xy = 3x^{3}y + 8x^{2}y^{2} + 4xy^{3} = xy(4y^{2} + 3x^{2} + 8xy)$
所以$\frac{16y^{2} + 9x^{2} + 14xy}{4y^{2} + 3x^{2} + 8xy} = xy = \frac{16\frac{y^{2}}{x^{2}} + 9 + 14\frac{y}{x}}{4\frac{y^{2}}{x^{2}} + 3 + 8\frac{y}{x}}$；
又因为$x,y$均是正数，令$\frac{y}{x} = t \in (0, + \infty)$，则$xy = f(t) = \frac{16t^{2} + 14t + 9}{4t^{2} + 8t + 3}$
所以， $f(t) = \frac{16t^{2} + 14t + 9}{4t^{2} + 8t + 3} = 4 - \frac{18t + 3}{4t^{2} + 8t + 3} = 4 - \frac{1}{\frac{4t^{2} + 8t + 3}{18t + 3}}$
令$g(t) = \frac{4t^{2} + 8t + 3}{18t + 3}$，
则$g(t) = \frac{2}{9}t + \frac{11}{27} + \frac{\frac{16}{9}}{18t + 3} = \frac{2}{9}\left( t + \frac{1}{6} \right) + \frac{\frac{16}{9}}{18t + 3} + \frac{10}{27} \geq 2\sqrt{\frac{2}{9}\left( t + \frac{1}{6} \right) \times \frac{\frac{16}{9}}{18t + 3}} + \frac{10}{27} = \frac{18}{27}$
当且仅当$\frac{2}{9}\left( t + \frac{1}{6} \right) = \frac{\frac{16}{9}}{18t + 3}$，即$t = \frac{1}{2}$时，等号成立；
所以$f(t) = 4 - \frac{1}{\frac{4t^{2} + 8t + 3}{18t + 3}} \geq 4 - \frac{1}{\frac{18}{27}} = \frac{45}{18} = \frac{5}{2}$
所以$f(t)$的最小值为$f{(t)}_{\min} = \frac{5}{2}$;
即当$t = \frac{y}{x} = \frac{1}{2},x = 2y = \sqrt{5}$时，即$x = \sqrt{5},y = \frac{\sqrt{5}}{2}$时，等号成立.
故答案为：$\frac{5}{2}$