Question

若a.b为有理数，且满足a^2/b^2<3,那么(a+3b)^2/(a+b)^2与3的大小关系是

Answer 1

3-(a+3b)^2/(a+b)^2=
=[3a^2+6ab+3b^2-(a^2+6ab+9b^2)]/(a+b)^2
=(2a^2-6b^2)/(a+b)^2
=2(a^2-3b^2)/(a+b)^2
因为a^2/b^2<3，(a+b)^2大于等于0，
所以3-(a+3b)^2/(a+b)^2=2(a^2-3b^2)/(a+b)^2<0
所以3<(a+3b)^2/(a+b)^2

Answer 2

(a+3b)^2/(a+b)^2=[(a+3b)/(a+b)]^2=[(a+b)/(a+b)+2b/(a+b)]^2=[1+2b/(a+b)]^2
a^2/b^2<3
则a/b<√3(若ab异号，则<0仍满足<√3)
2b/(a+b)=2/(a/b+1)
a/b+1<√3+1
2b/(a+b)=2/(a/b+1)>2/(√3+1)=√3-1
1+2b/(a+b)>1+√3-1=√3
[1+2b/(a+b)]^2>3
则(a+3b)^2/(a+b)^2=[1+2b/(a+b)]^2>3

Answer 3

3-(a+3b)^2/(a+b)^2
=[3(a+b)^2-(a^2+6ab+9b^2)]/(a+b)^2
=(2a^2-6b^2)/(a+b)^2
=2(a^2-3b^2)/(a+b)^2
=2(a^2/b^2-3)/(a/b+1)^2
因为，(a/b+1)^2>0 (分母不为0)
a^2/b^2<3 a^2/b^2-3<0
上式<0
所以3<(a+3b)^2/(a+b)^2