∞、若ab≠1,且有5a²+2001a+9=0及9b²+2001b+5=0,则a/b=

∞、若ab≠1,且有5a²+2001a+9=0及9b²+2001b+5=0,则a/b=
∞、若ab≠1,且有5a²+2001a+9=0及9b²+2001b+5=0,则a/b=

∞、若ab≠1,且有5a²+2001a+9=0及9b²+2001b+5=0,则a/b=
a=0时,等式5a²+2001a+9=0变为9=0,等式不成立,因此a≠0,同理,b≠0.
5a²+2001a+9=0 (1)
9b²+2001b+5=0 (2)
(1)×5-(2)×9
25a²+2001×5a-81b²-2001×9b=0
(25a²-81b²)+2001×(5a-9b)=0
(5a+9b)(5a-9b)+2001×(5a-9b)=0
(5a-9b)(5a+9b+2001)=0
(5a-9b)(5a²+9ab+2001a)=0 这一步是等式左边乘以a,等式仍成立.
5a²+2001a=-9代入,得(5a-9b)(9ab-9)=0
(5a-9b)(ab-1)=0
ab≠1,要等式成立,只有5a-9b=0
a/b=9/5