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[主观题]
证明:函数f(x)是n次多项式,a是方程f(x)=0的k(k≤m)重根f(a)=f´(a)==f(k-1)(a)=0,而f
证明:函数f(x)是n次多项式,a是方程f(x)=0的k(k≤m)重根f(a)=f´(a)==f(k-1)(a)=0,而f
证明:函数f(x)是n次多项式,a是方程f(x)=0的k(k≤m)重根f
(a)=f´(a)==f(k-1)(a)=0,而f(k)(a)≠0.
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