1.若bc-ad≥0,bd>0,求证(a+b)/b≥(c+d)/d
1个回答

这些题目都用作差法.

1.

(a+b)/b-(c+d)/d (通分)

=[(a+b)d-(c+d)b]/(bd)

=(ad-bc)/(bc) (ad-bc≥0,bc>0)

≥0

即 (a+b)/b≥(c+d)/d.

2.

a/(c-a)-b/(c-b) (通分)

=[a(c-b)-b(c-a)]/[(c-a)(c-b)]

=(ac-bc)/[(c-a)(c-b)]

=c(a-b)/[(c-a)(c-b)] (c>0,a-b>0,c-a>0,c-b>0)

>0

即 a/(c-a)>b/(c-b).

3.

(a+m)/(b+m)-a/b (通分)

=[(a+m)b-a(b+m)]/[b(b+m)]

=m(b-a)/[b(b+m)] (m>0,b-a>0,b>0,b+m>0)

>0

即 (a+m)/(b+m)>a/b.