Bài 2 : Theo ví dụ trên ta có : \(\frac{a}{b}< \frac{c}{d}\)=> ad < bc

Suy ra :

\(\Leftrightarrow ad+ab< bc+ba\Leftrightarrow a(b+d)< b(a+c)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)

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

\(\Leftrightarrow d(a+c)< (b+d)c\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\)

Vậy : ....

b, Theo câu a ta lần lượt có :

\(-\frac{1}{3}< -\frac{1}{4}\Rightarrow-\frac{1}{3}< -\frac{2}{7}< -\frac{1}{4}\)

\(-\frac{1}{3}< -\frac{2}{7}\Rightarrow-\frac{1}{3}< -\frac{3}{10}< -\frac{2}{7}\)

\(-\frac{1}{3}< -\frac{3}{10}\Rightarrow-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}\)

Vậy : \(-\frac{1}{3}< -\frac{4}{13}< -\frac{3}{10}< -\frac{2}{7}< -\frac{1}{4}\)

hoc24.net giúp em với

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

+) \(ad+ab< bc+ab\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\)( 1 )

+) \(ad+cd< bc+cd\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\Leftrightarrow\frac{a+c}{b+d}< \frac{c}{d}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\)

 Vì \(b,d>0\Rightarrow bd>0\)

\(\Rightarrow ad< bc\)

Ta lại có:

\(\frac{a}{b}=\frac{a\left(b+d\right)}{b\left(b+d\right)}=\frac{ab+ad}{b\left(b+d\right)}\)

\(\frac{a+c}{b+d}=\frac{b\left(a+c\right)}{b\left(b+d\right)}=\frac{ab+bc}{b\left(b+d\right)}\)

Vì \(b,d>0\)

Nên \(b\left(b+d\right)>0\)và \(d\left(b+d\right)>0\)         \(\left(1\right)\)

Mà \(ad< bc\Leftrightarrow ab+ad< ab+bc\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)ta có: \(\frac{ab+ad}{b\left(b+d\right)}>\frac{ab+bc}{b\left(b+d\right)}\)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(\cdot\right)\)

Ta lại có:

\(\frac{a+c}{b+d}=\frac{d\left(a+c\right)}{d\left(b+d\right)}=\frac{ad+cd}{d\left(b+d\right)}\)

\(\frac{c}{d}=\frac{c\left(b+d\right)}{d\left(b+d\right)}=\frac{bc+cd}{d\left(b+d\right)}\)

Mà \(ad< bc\Rightarrow ad+cd< bc+cd\left(3\right)\)

Từ \(\left(1\right)\)và \(\left(3\right)\)ta có:

\(\frac{ad+cd}{d\left(b+d\right)}< \frac{bc+cd}{d\left(b+d\right)}\)

\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(\cdot\cdot\right)\)

Từ \(\left(\cdot\right)\)và \(\left(\cdot\cdot\right)\)ta có: \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

\(\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)

\(\Rightarrow ab+ad< bc+ab\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)( 1 )

Lại có : ad < bc

\(\Rightarrow ad+cd< bc+cd\)

\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

Gỉa sử : \(\frac{a}{b}< \frac{a+c}{b+c}< =>ab+ac< ab+bc\)

\(< =>ac< bc< =>a< b\)(đpcm)

Gỉa sử : \(\frac{a}{b}>\frac{a+c}{b+c}< =>ab+ac>ab+bc\)

\(< =>ac>bc< =>a>b\)(đpcm)

ta có

a,\(\frac{a}{b}< 1\Leftrightarrow a< b\Leftrightarrow a+m< b+m\)

vì \(a+m< b+m\)

nên \(\frac{a+m}{b+m}< 1\)

b,Ta có    \(a+b>1\Leftrightarrow a+m>b+m\)

Vì \(a+m>b+m\)

nên \(\frac{a+m}{b+m}>1\)