Question

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

Answer 1

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

[2b-ab]^2-4*2[a-b][ab-2a]=0

4b^2+a^2b^-4ab^2-8[a^2b-2a^2-ab^2+2ab]=0

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

[2b+ab]^2-8a[2b+ab]+16a^2=0

[2b+ab-4a]^2=0

2b+ab-4a=0

2b-4a=-ab

2/a-4/b=-1

2/b-1/a=1/2