f(x)=2(x的3次方)—3x+1零点个数为多少个:
问题描述：
f(x)=2(x的3次方)—3x+1零点个数为多少个:
答
f(x)=2x³-2x²-x+1
=2x²(x-1)-(x-1)
=(2x²-1)(x-1)
=(√2x+1)(√2x-1)(x-1)
三个零点