∵|x?3|+|y+ 1 3 |＝0∴x-3=0；y+ 1 3 =0∴解得：x=3；y=- 1 3 ．原式=3x2y-（2xy-2xy+3x2y+x2y2）=3x2y-3x2y-x2y2=-x2y2当x=3；y=- 1 3 时，原式=?32×(? 1 3 )2=-1．
