作业帮f(x)=x(x-1)(2x-1)(3x-1)···(9x-1),则方程f(x)=0在[0,1]内有几个实根,f’(x)=0在（0,1）内有几个实根
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![f(x)=x(x-1)(2x-1)(3x-1)···(9x-1),则方程f(x)=0在[0,1]内有几个实根,f’(x)=0在（0,1）内有几个实根](/uploads/image/z/3768915-3-5.jpg?t=f%28x%29%3Dx%28x-1%29%282x-1%29%283x-1%29%C2%B7%C2%B7%C2%B7%289x-1%29%2C%E5%88%99%E6%96%B9%E7%A8%8Bf%28x%29%3D0%E5%9C%A8%5B0%2C1%5D%E5%86%85%E6%9C%89%E5%87%A0%E4%B8%AA%E5%AE%9E%E6%A0%B9%2Cf%E2%80%99%28x%29%3D0%E5%9C%A8%EF%BC%880%2C1%EF%BC%89%E5%86%85%E6%9C%89%E5%87%A0%E4%B8%AA%E5%AE%9E%E6%A0%B9)
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f(x)=x(x-1)(2x-1)(3x-1)···(9x-1),则方程f(x)=0在[0,1]内有几个实根,f’(x)=0在（0,1）内有几个实根
f(x)=x(x-1)(2x-1)(3x-1)···(9x-1),则方程f(x)=0在[0,1]内有几个实根,f’(x)=0在（0,1）内有几个实根
f(x)=x(x-1)(2x-1)(3x-1)···(9x-1),则方程f(x)=0在[0,1]内有几个实根,f’(x)=0在（0,1）内有几个实根
代入法：f(0)=0 f(1)=0 f(1/2)=0 f(1/3)=0.f(1/9)=0 你数数有几个 第二问：导数为0 所以两个实根之间导数有一个为零!你再数数!不懂再找我!
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