Question

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

Answer 1

（x＋2）²﹢|y﹢1|=0
所以
（x＋2）²=0,|y﹢1|=0
从而
x+2=0,y+1=0
x=-2,y=-1
所以
5xy²﹣﹛2x²y﹣[3xy²﹣﹙4xy²﹣2x²y﹚﹜
=5xy²﹣2x²y+[3xy²﹣﹙4xy²﹣2x²y﹚]
=5xy²﹣2x²y+3xy²﹣4xy²+2x²y
=4xy²
=4×(-2)×(-1)²
=-8

Answer 2

（x＋2）²﹢|y﹢1|=0
根据平方和根号的性质有：
x+2=0,y+1=0
x=-2,y=-1
对原代数式进行化简
5xy²﹣﹛2x²y﹣[3xy²﹣﹙4xy²﹣2x²y﹚]﹜
＝5xy²﹣2x²y+[3xy²﹣﹙4xy²﹣2x²y﹚]
＝5xy²﹣2x²y+3xy²﹣﹙4xy²﹣2x²y﹚
＝5xy²﹣2x²y+3xy²﹣4xy²+2x²y
＝4xy²
把x,y代入得
=4*(-2)*(-1)^2
==-8

Answer 3

前面是零加零形式，解得x=-2,y=-1
5xy²﹣﹛2x²y﹣[3xy²﹣﹙4xy²﹣2x²y﹚﹜=-8