解不等式x³-3x²+x+1<0
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解不等式x³-3x²+x+1<0
借助啊.
借助啊.
![解不等式x³-3x²+x+1<0](/uploads/image/z/6490888-16-8.jpg?t=%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Fx%26%23179%3B%EF%BC%8D3x%26%23178%3B%EF%BC%8Bx%EF%BC%8B1%EF%BC%9C0)
x³-3x²+x+1
=x³-2x²+x-x²+1
=x(x-1)²-(x-1)(x+1)
=(x-1)[x(x-1)-(x+1)]
=(x-1)(x²-2x-1)
当x-1>0,即x>1时,
x²-2x-1
=x³-2x²+x-x²+1
=x(x-1)²-(x-1)(x+1)
=(x-1)[x(x-1)-(x+1)]
=(x-1)(x²-2x-1)
当x-1>0,即x>1时,
x²-2x-1