Question

单项式3x^2y,-3xy^2,2ⅹ^2y,2xy^2,-1和是多少?

先化简,再求值：
3x^2y-[2xy^2-2（xy-3/2x^2y）+xy]+3xy^2,其中x=8,y=-1/2

Answer 1

原式=3x^2y-(2xy^2-2xy+3x^2y+xy)+3xy^2
=3x^2y-(2xy^2-xy+3x^2y)+3xy^2
=3x^2y-2xy^2+xy-3x^2y+3xy^2
=xy^2+xy
=8*1/4+8*(-1/2)
=-2