已知：（x+2)²+丨y-6丨=0,求2（xy²+x²y)-[2xy²-3(1-x²y)]-2的值.

题目

已知：（x+2)²+丨y-6丨=0,求2（xy²+x²y)-[2xy²-3(1-x²y)]-2的值.

答案

（x+2)²+丨y-6丨=0
∴﹛x+2=0
y-6=0
∴x=-2,y=6
2（xy²+x²y)-[2xy²-3(1-x²y)]-2
=2xy²+2x²y-(2xy²-3+3x²y)-2
=2xy²+2x²y-2xy²+3-3x²y-2
=-x²y+1
=-(-2)²×6+1
=-23