Question

已知（x+2)²+|y+1|=0,求5xy²-{2x²y-[3xy²-(4xy²-2x²y)]}的值

sup2=平方

Answer 1

5xy²御老-{2x²y-[3xy²-(4xy²-2x²y)]}
=5xy²-{2x²y-[3xy²-4xy²+2x²y]}
=5xy²-{2x²y-3xy²+4xy²-2x²y}
=5xy²-2x²y+3xy²-4xy²+2x²y
=4xy²物拆腔
因（x+2)²+|y+1|=0，则（x+2)²=0，|y+1|=0
所以x=-2,y=-1
带入原罩衫式=4*（-2）*（-1）²=-8

Answer 2

5xy²-{2x²y-[3xy²-(4xy²-2x²y)]}
=5xy²-{2x²y-[3xy²-(4xy²-2x²y)]}
=5xy²-{2x²y-[3xy²-4xy²+2x²y]}
=5xy²-{2x²y-3xy²+4xy²-2x²y}
=5xy²-2x²y+3xy²-4xy²+2x²y
=4xy²
因为（罩衫x+2)²+|y+1|=0，所以（x+2)²，|y+1是二个非负数
所以物拆腔x=-2,y=-1
带入，得
原式=4*（-2）*（-1）御老²=-8