x²/4-y²=1，则3x²-2xy=(3x²-2xy)/(x²/4-y²)=(12x²-8xy)/(x²-4y²)=[12(x/y)²-8(x/y)]/[(x/y)²-4].设x/y=z，且上式等于t，则(12z²-8z)/(z²-4)=t，即(t-12)z²+8z-4t=0.∴△=64+16t(t-12)≥0解得，t≥6+4√2或t≤6-4√2，故所求最小值为6+4√2;所求最大值为6-4√2。
