赞 新地123 春芽
共回答了21个问题采纳率：95.2% 举报
(1):
f(-x)=(-x+a)/(x^2-bx+1) = -（x - a)/(x^2-bx+1)
-f(-x) = f(x)
（x - a)/(x^2-bx+1) = (x+a)/(x^2+bx+1)
（x - a)(x^2+bx+1) = (x+a)(x^2-bx+1)
x^3 +bx^2 +x - ax^2 -abx -a = x^3 -bx^2 + x + ax^2 -abx +a
2bx^2 - 2ax^2 - 2a = 0
(b - a)x^2 = a
x取任何值时此式都成立,那么只有：
b-a = 0
a = 0
即a = b = 0
(2):
f(x) = x/(x^2 + 1) = 1/(x + 1/x)
考虑x>0时的情况
[√x - √（1/x)]^2 >= 0
x + 1/x - 2 >= 0
x + 1/x >= 2
f(x) = 1/(x + 1/x)
f(-x)=(-x+a)/(x^2-bx+1) = -（x - a)/(x^2-bx+1)
-f(-x) = f(x)
（x - a)/(x^2-bx+1) = (x+a)/(x^2+bx+1)
（x - a)(x^2+bx+1) = (x+a)(x^2-bx+1)
x^3 +bx^2 +x - ax^2 -abx -a = x^3 -bx^2 + x + ax^2 -abx +a
2bx^2 - 2ax^2 - 2a = 0
(b - a)x^2 = a
x取任何值时此式都成立,那么只有：
b-a = 0
a = 0
即a = b = 0
(2):
f(x) = x/(x^2 + 1) = 1/(x + 1/x)
考虑x>0时的情况
[√x - √（1/x)]^2 >= 0
x + 1/x - 2 >= 0
x + 1/x >= 2
f(x) = 1/(x + 1/x)
1年前
6
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