2x^2y-(5xy^2-4(xy^2-4分之3x二次方y)+x二次方y)+3xy二次方,其中x,y满足:|x-1|+(y+2)二次方=0
问题描述:
2x^2y-(5xy^2-4(xy^2-4分之3x二次方y)+x二次方y)+3xy二次方,其中x,y满足:|x-1|+(y+2)二次方=0
答
x,y满足:|x-1|+(y+2)二次方=0
那么x-1=0,y+2=0
所以x=1,y=-2
所以2x^2y-(5xy^2-4(xy^2-4分之3x二次方y)+x二次方y)+3xy二次方
=2x^2y-(5xy^2-4xy^2+3x^2y+x^2y)+3xy^2
=2x^2y-5xy^2+4xy^2-3x^2y-x^2y+3xy^2
=-2x^2y+2xy^2
=2xy(-x+y)
=2*1*(-2)*(-1-2)
=12
如果不懂,祝学习愉快!