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) 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

So sánh: \(\dfrac{434}{561}\) và \(\dfrac{441}{568}\)

* Bài làm:

Vì \(\dfrac{434}{561}\) < 1 => \(\dfrac{434}{561}\) < \(\dfrac{434+7}{561+7}\) hay \(\dfrac{434}{561}\) < \(\dfrac{441}{568}\)

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

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

\(\dfrac{a}{b}\) < \(1\) \(\Rightarrow\) \(a\) < \(b\), suy ra \(ab+am\) < \(ab+bm\). (3)

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

b) Áp dụng, rõ ràng \(\dfrac{434}{561}\) < 1 nên \(\dfrac{434}{561}\) < \(\dfrac{434+7}{561+7}\)=\(\dfrac{441}{568}\)