Question

设b＜a＜0，a^2+b^2=5/2ab,则(a+b)/(a-b)等于

Answer 1

方法很多：∵a^2+b^2=5ab/2,
∴2a^2-5ab+2b^2=0,
即(2a-b)(a-2b)=0
b=2a,或a=2b
∵b＜a＜0,∴b-a＜0,
则a=2b,不满足条件，舍去，只去b=2a
∴(a+b)/(a-b)=3a/(-a）=-3
法二：(a+b)^2/(a-b)^2
=(a^2+b^2+2ab)/(a^2+b^2-2ab)
=(5ab/2+2ab)/(5ab/2-2ab)=9
∵b＜a＜0∴a+b＜0,a-b＞0
∴(a+b)/(a-b)=-3

Answer 2

a^2+b^2=5/2ab
=> a^2+b^2 + 2ab =5/2ab +2ab => (a+b)^2 = 4.5ab
=> a^2+b^2 - 2ab =5/2ab -2ab => (a-b)^2 = 0.5ab
所以[(a+b)/(a-b)]^2=9
又因为b<a<0,所以a+b<0,a-b>0
因此(a+b)/(a-b)=-3

Answer 3

由a^2+b^2=5ab/2，可得：
(a+b)^2=9ab/2
(a-b)^2=ab/2
又因b＜a＜0，所以a+b<0,a-b>0
所以
a+b=-3√ab/√2
a-b=√ab/√2
所以a+b)/(a-b)=-3