a+m/b+m > a/b

ta xét 3 trường hợp\(\frac{a}{b}\)= 1 ; \(\frac{a}{b}\)< 1 ; \(\frac{a}{b}\)> 1

+ trương hợp \(\frac{a}{b}\)= 1 nên a = b thì \(\frac{a+b}{b+m}\)= \(\frac{a}{b}\)= 1

+ trường hợp \(\frac{a}{b}\)< 1 nên a < b nên a + b < b + m

còn lại tự làm nhé

                  Giải

Xét 3 tường hợp : \(\frac{a}{b}=1;\frac{a}{b}>1;\frac{a}{b}< 1\)

\(TH1:\frac{a}{b}=1\Leftrightarrow a=b\)

\(\Rightarrow\frac{a+m}{b+m}=\frac{b\left(a=b\right)+m}{b+m}=1\)

\(\Rightarrow\frac{a+m}{b+m}=\frac{a}{b}\)

\(TH2:\frac{a}{b}>1\Leftrightarrow a>b\)

Ta có : \(b\left(a+m\right)< a\left(b+m\right)\) ( tích chéo )

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

\(\Leftrightarrow bm< am\)( luôn đúng )

\(\Rightarrow\frac{a+m}{b+m}< \frac{a}{b}\)

\(TH3:\frac{a}{b}< 1\Leftrightarrow a< b\)

Ta có : \(b\left(a+m\right)>a\left(b+m\right)\) ( tích chéo )

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

​\(\Leftrightarrow bm>am\)​( luôn đúng )

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

TH1 : a<b

\(\Rightarrow am< bm\)

\(\Rightarrow ab+am< ab+bm\Rightarrow a\left(b+m\right)< b\left(a+m\right)\)

\(\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\)

TH2 : a=b

\(\Rightarrow am=bm\)

\(\Rightarrow ab+am=ab+bm\Rightarrow a\left(b+m\right)=b\left(a+m\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a+m}{b+m}\)

TH1 : a>b

\(\Rightarrow am>bm\)

\(\Rightarrow ab+am>ab+bm\Rightarrow a\left(b+m\right)>b\left(a+m\right)\)

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

Vậy ... ( có 3 trường hợp )

a/b lon hon k nha

a + m b + m = a + m . b b + m . b = a b + m b b + m . b ; a b = a b + m b + m . b = a b + a m b + m . b

Do  a > b ⇒ a m > b m ⇒ a b + m b > a b + a m ⇒ a + m b + m > a b