Question

已知x²+x-1=0求x³+2x²+2012的值 已知x²-5x=6,求（x-1

Answer 1

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