已知xy=35,求x2−xy+2y2x2+2xy−y2的值.

已知xy=35,求x2−xy+2y2x2+2xy−y2的值.

∵xy=35,∴yx=53,x2−xy+2y2x2+2xy−y2分子分母同時除以xy得,xy−1+2yxxy+2−yx,∴原式=xy−1+2yxxy+2−yx=35−1+10335+2−53=44151415=4414=227.