﹣2x-﹛4x-2y-[3x-﹙2y+1﹚]﹜,其中x=﹣3分之2,y=2008
问题描述:
﹣2x-﹛4x-2y-[3x-﹙2y+1﹚]﹜,其中x=﹣3分之2,y=2008
答
原式=-2x-{4x-2y-[3x-(2y+1)]}
=-2x-[4x-2y-(3x-2y-1)]
=-2x-(4x-2y-3x+2y+1)
=-2x-4x+2y+3x-2y-1
=(-2x-4x+3x)+(2y-2y)-1
=-3x-1
当x=-2/3,y=2008时
原式=-3×(-2/3)-1
=2-1
=1