已知（x+2）∧2+y+1的绝对值＝0,求5xy∧2-{2x∧2y-【3xy^2】-（4xy^2-2x∧2y）}的值

根据已知 得到 x=-2 y=-1
原式=5xy^2 - {2x^2y-[3xy^2-(4xy^2-2x^2y)]}
=5xy^2 - {2x^2y-[3xy^2-4xy^2+2x^2y]}
=5xy^2 - {2x^2y-[-xy^2+2x^2y]}
=5xy^2 - {2x^2y+xy^2-2x^2y]}
=5xy^2 - xy^2
=4xy^2
=-8