1、x³+2x²+2012=x(x²+x-1)+(x²+x-1)+1+2012=2013
2、x²-5x=6,则(x-2)(x-3)=12；(x-1)(x-4)=10；(x-1)(x-2)(x-3)(x-4)=120.
3、-xy[x³(y^7)-3x²(y^5)-5y]=-xy[(xy²)³y-(xy²)²3y-5y]=-xy(-216y-108y-5y)=-xy(-329y)=329xy²=-1974第二天题是......120还是0啊当然是120。x²-5x+6=(x-2)(x-3)=12；x²-5x+4=(x-1)(x-4)=10。相乘得0。若x²-5x=-6，x²-5x+6=0，(x-2)(x-3)=0，(x-1)(x-2)(x-3)(x-4)=0，很明显0是错的。Ŷ�ðɣ���֪�����а�������ɺðɣ�