Question

已知（x+2)^2+|y+1|=0,求5xy^2-{2x^2y-[3xy^2-(4xy^2-2x^2y)]}

Answer 1

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

Answer 2

因为一个数的平方大于等于0
一个数的相反数也大于等于0
所以只有当 x + 2 = 0 且 y + 1 = 0 时等号才成立
所以 x = -2 , y = -1

5xy² - {2x²y -[3xy² -(4xy² -2x²y)]}
= 5xy² - 2x²y + 3xy² - 4xy² + 2x²y
= 4xy²
= 4×(-2)×(-1)²
= -8