Ta có a<b 

=>ac<bc (c>0)

=> ac+ ab < bc+ ab

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

=> a/b< a+c/b+c(đpc/m)

\(\frac{a}{b}=\frac{ab+a}{b^2+b};\frac{a+1}{b+1}=\frac{ab+b}{b^2+b}\)

\(+,a>b\Rightarrow ab+a>ab+b\Rightarrow\frac{a}{b}>\frac{a+1}{b+1}\left(vì:b>0\right)\)

\(+,a=b\Rightarrow\frac{a}{b}=\frac{a+1}{b+1}=1\)

\(+,a< b\Rightarrow ab+a< ab+b\Rightarrow\frac{a}{b}< \frac{a+1}{b+1}\left(vì:b>0\right)\)

\(Vậy:voi:a>b\text{ thì }\frac{a}{b}>\frac{a+1}{b+1};voi:a=b\text{ thì: }\frac{a}{b}=\frac{a+1}{b+1}=1;voi:a< b\text{ thì:}\frac{a}{b}< \frac{a+1}{b+1}\)

C1 : Theo ví dụ trên ta có : \(\frac{a}{b}< \frac{c}{d}\)=> ad < bc

Suy ra :

<=> ad + ab < bc + ba <=> a[b + d] < b[a + c] <=> \(\frac{a}{b}< \frac{a+c}{b+d}\)

Mặt khác ad  < bc => ad + cd < bc + cd

<=> d[a + c] < [b + d]c <=> \(\frac{a+c}{b+d}< \frac{c}{d}\)

Từ đó suy ra \(\frac{a}{b}< \frac{a+c}{b+c}< \frac{c}{d}\)

C2 : Xét hiệu : \(\frac{a+c}{b+d}-\frac{a}{b}=\frac{ab+bc-ab-ad}{b(b+d)}=\frac{bc-ad}{b(b+d)}>0\)

\(\frac{c}{d}-\frac{a+c}{b+d}=\frac{bc+cd-ad-cd}{d(b+d)}=\frac{bc-ad}{d(b+d)}>0\)

giup My vs

\(\frac{a}{b}< \frac{c}{d}\) => ad < bc

=> ad + ab < bc + ab

=> a(b + d) < b(a + c)

=> \(\frac{a}{b}< \frac{a+c}{b+d}\)

=> ad < bc

=> ad + cd< bc + cd

=> d(a + c) < c(b + d)

=> \(\frac{a+c}{b+d}< \frac{c}{d}\)

=> đccm

b) \(\frac{-1}{3}=\frac{-16}{48}< \frac{-15}{48}\); \(\frac{-14}{48};\frac{-13}{48}\)\(< \frac{-12}{48}=\frac{-1}{4}\)

ok mk nhé!!! 4556577568797902451353466545475678769863513532345634645645745

ghi cho minh de hieu nha