2x^2-4x+1=0,(x+1)(x+2)(x-3)(x-4)=?
问题描述:
2x^2-4x+1=0,(x+1)(x+2)(x-3)(x-4)=?
答
2x²-4x=-1
x²-2x=-1/2
原式=[(x+1)(x-3)][(x+2)(x-4)]
=(x²-2x-3)(x²-2x-8)
=(-1/2-3)(-1/2-8)
=119/4