若函数f（x）=x³-ax（a大于零）都在区间【-10,10】上,则使得方程f（x）=1000有正整数解的实数a的取值个数x³-ax-1000=0,故得a=(x³-1000)/x,x∈[-10,10].不难求得,当x=-10,-8,..-5,-4,-2,..-1,1,2,4,8,...