设函数f(x)=(x2-10x+c1)(x2-10x+c2)(x2-10x+c3)(x2-10x+c4)(x2-10x+c5),设集合M={x|f(x)=0}={x1,x2,...,x9}⊆N+,
设c1≥c2≥c3≥c4≥c5 ,则c1-c5为( )
若b,c∈[-1,1],则方程x2+2bx+c=0有实数根的概率为
A.B.C.D.
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