suppose you are right

suppose you are right,

since p,q, r are logically symetric,

if you can proof b=3+p,a=2/(6-3p)

then you can also proof b=3+q,a=2/(6-3q)

and b=3+r,a=2/(6-3r)

so you get p=q=r (you know why?) ====> something may be wrong in your calcul!!!

--------------------------------------------

I do some calculs also, f(5) = 3a(5-b), a root of f(x), you get

1) 3a(5-b) = b, or

2) 3a(5-b) = 2

in both case, you replace a by 2/(6-3p), b by 3+p,you can find p

but you said "p can take any complex number,if one of q,r is not real number"

Absurd! so you are not right

所有跟帖： 

• great! -15少- ♂ (391 bytes) () 07/16/2011 postreply 17:08:12

• Sir,b is comlex number,you are too... -jinjing- ♀ (97 bytes) () 07/16/2011 postreply 18:57:01

• Let's skip this...... though a little competition is good. Jing -乱弹- ♂ (0 bytes) () 07/17/2011 postreply 18:07:29

• (cont), maybe you can post some new problems. -乱弹- ♂ (0 bytes) () 07/17/2011 postreply 18:08:01

• yes, let's skip this...... it is becoming boring -15少- ♂ (0 bytes) () 07/18/2011 postreply 05:39:40

• 回复：suppose you are right -jinjing- ♀ (35 bytes) () 07/16/2011 postreply 19:47:45

• You are mystifying ... -15少- ♂ (1077 bytes) () 07/17/2011 postreply 02:41:26

• I'm sorry ,I said complex,is real complex, means not real. -jinjing- ♀ (356 bytes) () 07/19/2011 postreply 06:35:33