设f(x) = ax² + bx + cf(x) ≥ a+1-xax² + bx + c ≥ a+1-xax² + (b+1)x + c-a-1 ≥ 0解集为[-1,1],则ax² + (b+1)x + c-a-1 ≥ 0,g(x) = ax² + (b+1)x + c-a-1可表达为a(x + 1)(x - 1) 而...