✔ 最佳答案
(x + a)(x - b) = 0
a) False
a + b = 0, ie. a = -b
(x + a)(x - b) = (x - b)(x - b) = (x - b)^2 >/= 0
(greater than or equal to 0)
b) False
x + a = 0 & x - b = 0, ie. x = -a & x = b
Let (x1 + a)(x2 - b) = 0,
If x1 = -a, x2 = b, (x1 + a)(x2 - b) = (-a + a)(b - b) = 0*0 = 0
If x1 = b, x2 = -a, (x1 + a)(x2 - b) = (b + a)(-a - b) = -(a + b)^2 </= 0
( smaller than or equal to 0)
c) True
x + a = 0 or x - b = 0, ie. x = -a or x = b
If x = -a, (x + a) = 0 => (x + a)(x - b) = 0
If x = b, (x - b) = 0 => (x + a)(x - b) = 0
=> (x + a)(x - b) = 0
d) False, as it can be greater than/ smaller than/ equal to 0
a x b = 0, ie. a = 0 or b = 0
Consider a = 0, (x + a)(x - b) = x(x - b)
When x > b, x >/= 0
x(x - b) >/= 0
You can prove others, eg.
When x > b, x < 0
x(x - b) < 0
e) False
None of the above
2009-04-08 13:55:39 補充:
d) when a = b = 0,
(x + a)(x - b) = x^2 >/= 0