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设f(x)=anxn+an-1xn-1+…+a1x+a0是实系数多项式,n≥2,且某个ak=0(1≤k≤n-1),及当i≠k时,ai≠0。证明:若f(x)有n个
设f(x)=anxn+an-1xn-1+…+a1x+a0是实系数多项式,n≥2,且某个ak=0(1≤k≤n-1),及当i≠k时,ai≠0。证明:若f(x)有n个相异的实根,则ak-1·ak+1<0
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设f(x)=anxn+an-1xn-1+…+a1x+a0是实系数多项式,n≥2,且某个ak=0(1≤k≤n-1),及当i≠k时,ai≠0。证明:若f(x)有n个相异的实根,则ak-1·ak+1<0