求初二一道数学题～会的，高分

a^2+2ab+b^2=6ab
(a+b)^2=6ab

a^2-2ab+b^2=2ab
(a-b)^2=2ab

((a+b)/(a-b))^2=(6ab)/(2ab)=3
a<b<0,所以a+b<0,a-b<0
(a+b)/(a-b)>0
(a+b)/(a-b)=根号3

根号3

(a+b)^2=a^2+b^2+2ab=6ab
(a-b)^2=a^2+b^2-2ab=2ab
因为a<b<0 a-b<0 ab>0
a-b=-根号2
a+b=-根号6
答案为 根号3

已知:a<b<0;a^2+b^2=4ab;那么(a+b)^2=a^2+b^2+2ab=4ab+2ab=6ab (1);(a-b)^2=a^2+b^2-2ab=4ab-2ab=2ab (2);两式相除得[(a+b)/(a-b)]^2=6ab/2ab=3;由知条件得a+b<0,a-b<0,a+b/a-b=3^1/2>0

(a+b)²+2ab=4ab (a+b)²=2ab
(a-b)²+6ab=4ab (a-b)²=-2ab
(a+b)²/(a-b)²=a+b/a-b= -1