证明方程x³+4x²_3x_1=0有三个实根
人气：333 ℃ 时间：2020-05-20 06:08:46
解答
设f(x)=x³+4x²-3x-1
f(0)=-10
因此在0与1之间必有一个0点
同理f(-1)=5>0
所以在-1与0之间必有一个0点
又f（-5）=-11
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