已知（x+2）²+│y-1│=0,求7x²y-3+2xy²-6x²y-2xy²+4的值

已知（x+2）²+│y-1│=0,求7x²y-3+2xy²-6x²y-2xy²+4的值
已知（x+2）²+│y-1│=0,求7x²y-3+2xy²-6x²y-2xy²+4的值

已知（x+2）²+│y-1│=0,求7x²y-3+2xy²-6x²y-2xy²+4的值
x=-2,y=1
7x²y-3+2xy²-6x²y-2xy²+4=5

(x+2)²+|y-1|=0 两个非负数的和等于0，这两个非负数都等于0
x+2=0 且 y-1=0
x=-2 , y=1
原式=(7x²y-6x²y)+(2xy²-2xy²)+(4-3)
=x²y+1
=(-2)²×1+1
=4×1+1
=5