Question

先化简，再求值：3x2y?[2xy?2(xy?32x2y)+x2y2]，其中|x?3|+|y+13|＝0

先化简，再求值：3x2y?[2xy?2(xy?32x2y)+x2y2]，其中|x?3|+|y+13|＝0．

Answer 1

∵|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．