a ) Nếu \(\frac{a}{b}>\frac{a+m}{b+m}\)

\(\Leftrightarrow a\left(b+m\right)>b\left(a+m\right)\)

\(\Leftrightarrow ab+am>ab+bm\)

\(\Leftrightarrow am>bm\)

\(\Rightarrow a>b\)

\(\Rightarrow\frac{a}{b}>1\)

Vậy \(\frac{a}{b}>1\) thì \(\frac{a}{b}>\frac{a+m}{b+m}\)

b ) Vì 237 > 142 => \(\frac{237}{142}>\frac{237+9}{142+9}=\frac{246}{151}\)

Xét hiệu :

\(\frac{a}{b}-\frac{a+m}{b+m}\)

\(=\frac{a\left(b+m\right)}{b\left(b+m\right)}-\frac{\left(a+m\right)b}{\left(b+m\right)b}\)

\(=\frac{a.b+a.m}{b\left(b+m\right)}-\frac{a.b+b.m}{b\left(b+m\right)}\)

\(=\frac{a.b+a.m-a.b+b.m}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}\)

Vì \(\frac{a}{b}>1,b\in\)N* \(\Rightarrow a>b\Rightarrow a-b>0,m\in\)N*

\(\Rightarrow m\left(a-b\right)>0\); Vì : \(b,m\in\)N* \(\Rightarrow b\left(b+m\right)>0\)

\(\Rightarrow\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\) hay : \(\frac{a}{b}-\frac{a+m}{b+m}>0\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\)

Vậy \(\frac{a}{b}>1,m\in\)N* thì \(\frac{a}{b}>\frac{a+m}{b+m}\)

b, Tự làm 

a. Ta có

\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}.\)

Vì\(\frac{2011}{2012+2013}< \frac{2011}{2012}.\)(1)

\(\frac{2012}{2012+2013}< \frac{2012}{2013}.\)(2)

Cộng vế với vế của 1;2 ta được

\(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}< A=\frac{2011}{2012}+\frac{2012}{2013}\)

hay A>B

Làm ơn giúp mk, mk đang cần gấp!!!

a.Ta có 

\(1-\frac{a}{b}=\frac{b-a}{b}.\)

\(1-\frac{a+m}{b+m}=\frac{a-b}{b+m}\)

Do \(m\in N;m\ne0\)nên \(\frac{b-a}{b}>\frac{b-a}{b+m}\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

hay: \(\frac{a}{b}< \frac{a+m}{b+m}\)

a) Vì a/b > 1 nên a > b

 Ta có: \(\frac{a}{b}-\frac{a+m}{b+m}=\frac{a\left(b+m\right)-b\left(a+m\right)}{b\left(b+m\right)}=\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\)

=> \(\frac{a}{b}>\frac{a+m}{b+m}\)

b) lấy a=237, b= 142; m = 9

\(\frac{237}{142}>\frac{237+9}{142+9}\)

mệt rồi mình nghỉ đây tối mình giải cho

1,\(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b^2+bm}\)

2,\(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b^2+bm}\)

3,\(\frac{a}{b}<1\) =>a<b =>ab+am<ab+bm

Từ 1,2 và 3 ta có :\(\frac{a}{b}<\frac{a+m}{b+m}\)

\(\frac{a}{b}\)< 1 <=> a < b <=> a.m < b.m <=> ab + a.m < ab + b.m

                                                       <=> a(b + m) < b(a + m)

                                                       <=> \(\frac{a}{b}\)< \(\frac{a+m}{b+m}\)