f(x)=a(x-2)(x-b)
f(f(x))=a[a(x-2)(x-B)-2][a(x-2)(x-b)-b)]
f(f(5))=a[a*3*(5-b)-2)][a*3(5-b)-b)]=0,As only 5 one real solution,in real number field,
We have:[a*3*(5-b)-2]=[a*3*(5-b)-b)]=0
We get b=2,a=2/9,so,f(0)=2/9(0-2)(0-2)=8/9
f(x)=a(x-2)(x-b)
f(f(x))=a[a(x-2)(x-B)-2][a(x-2)(x-b)-b)]
f(f(5))=a[a*3*(5-b)-2)][a*3(5-b)-b)]=0,As only 5 one real solution,in real number field,
We have:[a*3*(5-b)-2]=[a*3*(5-b)-b)]=0
We get b=2,a=2/9,so,f(0)=2/9(0-2)(0-2)=8/9
• 回复:回复:回复:回复:有没有好办法? -cma- ♂ (111 bytes) () 07/11/2011 postreply 15:21:06
• If they are not both 0, 5 is not only one root. -jinjing- ♀ (0 bytes) () 07/12/2011 postreply 07:11:08
• [a*3*(5-b)-2]=0 or [a*3*(5-b)-b)]=0.It can't get answer.Only bot -jinjing- ♀ (61 bytes) () 07/12/2011 postreply 10:37:39
• 5 -乱弹- ♂ (0 bytes) () 07/12/2011 postreply 20:13:46
• 愿闻其详. -jinjing- ♀ (0 bytes) () 07/13/2011 postreply 05:18:53
• 不知道答案,但可列两个方程。 -乱弹- ♂ (0 bytes) () 07/12/2011 postreply 20:17:43