Question

已知2(a-b)x^2+(2b-ab)x+(ab-2a)=0有两个相等的实根,则2/b-1/a=?

Answer 1

(2b-ab)^2-4*2(a-b)*(ab-2a）=0

(ab)^2-8a^b+16a^2-16ab+4ab^2+4b2=0

同时除 (ab)^2

1-8/b+16/b^2-16/(ab)+4/a+4/a^2=0

1-4*(2/b-1/a)+4*(2/b-1/a)^2=0

(2/b-1/a-1/4)^2=0

2/b-1/a=1/4