（5分）已知甲、乙两个圆台上下底面的半径均为$r_{2}$和$r_{1}$，母线长分别为$2(r_{1}-r_{2})$和$3(r_{1}-r_{2})$，则两个圆台的体积之比$\dfrac{{{V}_{}}}{{{V}_{}}}=$____．
分析：由已知结合圆台的体积公式即可求解．
解：因为甲、乙两个圆台上下底面的半径均为$r_{2}$和$r_{1}$，母线长分别为$2(r_{1}-r_{2})$和$3(r_{1}-r_{2})$，
则两个圆台的体积之比$\dfrac{{{V}_{}}}{{{V}_{}}}=\dfrac{\dfrac{1}{3}\left( {{S}_{1}}+{{S}_{2}}+\sqrt{{{S}_{1}}{{S}_{2}}} \right){{h}_{}}}{\dfrac{1}{3}\left( {{S}_{1}}+{{S}_{2}}+\sqrt{{{S}_{1}}{{S}_{2}}} \right){{h}_{}}}=\dfrac{{{h}_{}}}{{{h}_{}}}=\dfrac{\sqrt{{{[2\left( {{r}_{1}}-{{r}_{2}} \right)]}^{2}}-{{({{r}_{1}}-{{r}_{2}})}^{2}}}}{\sqrt{{{[3\left( {{r}_{1}}-{{r}_{2}} \right)]}^{2}}-{{({{r}_{1}}-{{r}_{2}})}^{2}}}}\dfrac{\sqrt{3}\left( {{r}_{1}}-{{r}_{2}} \right)}{2\sqrt{2}\left( {{r}_{1}}-{{r}_{2}} \right)}=\dfrac{\sqrt{6}}{4}$．
故答案为：$\dfrac{\sqrt{6}}{4}$．
点评：本题主要考查了圆台的体积公式的应用，属于基础题．