Question

已知(x+2)²+|y+1|=0,求5xy²-2x²y-[3xy²-(4xy²-2x²y)] 急急急！！谢谢

Answer 1

(x+2)²+|y+1|=0 两个非负数的和为零 他们都是零
x+2=0 y+1=0
x=-2 y=-1
5xy²-2x²y-[3xy²-(4xy²-2x²y)]
=5xy²-2x²y-[3xy²-4xy²+2x²y]
=5xy²-2x²y-3xy²+4xy²-2x²y
=5xy²-3xy²+4xy²-2x²y-2x²y
=(5-3+4)xy²-4x²y
=6xy²-4x²y
=2xy(3y-2x)
=2×2×(-3+4)
=4

Answer 2

(x+2)²+|y+1|=0
(x+2)²>=0,|y+1|>=0
(x+2)²=0,,|y+1|=0
x= -2,y= -1
5xy²-2x²y-[3xy²-(4xy²-2x²y)]
=5xy²-2x²y-(-xy²-2x²y)
=6xy²= -12

Answer 3

你好，很高兴为你解答~
通过题意知(x+2)²=0
|y+1|=0
所以得出X=-2 Y=-1
然后再化简5xy²-2x²y-[3xy²-(4xy²-2x²y)]
原式=5xy²-2x²y-[3xy²-4xy²+2x²y]
=5xy²-2x²y-[-xy²+2x²y]
=5xy²-2x²y+xy²-2x²y
=5xy²+xy²-2x²y-2x²y
=6xy²-4x²y
=xy（6y-4x）
代入X=-2 Y=-1得原式=（-2）*（-1）[(6*-1)-(4*-2)]
=2*[-6-(-8)]
=2*(-6+8)
=2*2
=4

希望能够帮助到你，谢谢~