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

\(\Rightarrow ad< bc\)

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

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

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

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

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Theo tính chất dãy tỉ số bằng nhau có:

\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)

\(\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

ta có a+b/a-b=c+d/c-d

suy ra (a+b)(c-d)=(a-b)(c+d)

ac-ad+bc-bd=ac+ad-bc-bd

ac-ac+bc+bc-bd+bd=ad+ad

2bc=2ad 

nen bc=ad=a/b=c/d

vay tu a/b=c/d ta co the suy ra a+b/a-b=c+d/c-d

a) \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (quy đồng mẫu chung)

Vì b,d > 0 nên bd > 0. Do đó ad < bc (đpcm)

b) ad < bc \(\Leftrightarrow\frac{ad}{bd}< \frac{bc}{bd}\) (cùng chia cho bd)

Vì b,d > 0 nên bd > 0. Do đó \(\frac{a}{b}< \frac{c}{d}\) (rút gọn tử và mẫu)

a, Ta có: \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{cb}{db}\Rightarrow ad< cb\) 

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

a=?

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

=> a²+ab+ad+db=cb+c²+db+dc

=> a²+ab+ad+db-cb-c²-db-dc=0

=>( a²-c²) + (ab -bc) +( ad -dc)=0

=>(a+c)(a-c) +b(a-c) +d(a-c)=0

=>(a-c)(a+c+b+d)=0

=>\(\left[\begin{array}{nghiempt}a-c=0\\a+b+c+d=0\end{array}\right.\)

=>\(\left[\begin{array}{nghiempt}a=c\\a+b+c+d=0\end{array}\right.\)

đề như đệt

a. Nếu : \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{a}{b}\times bd< \frac{c}{d}\times bd\left(\text{ do }bd>0\right)\)

\(\Leftrightarrow ad< bc\) vậy ta có điều phải chứng minh

b. nếu \(ad< bc\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Leftrightarrow\frac{a}{b}< \frac{c}{d}\) vậy ta có đpcm

Mk cảm ơn

bn vào câu hỏi tương tự

có người làm câu này rồi

a/b<c/d

=>ad<bc

=>ad+ab<bc+ab

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

=>a/b<a+c/b+d(1)

từ ad<bc

=>ad+cd<bc+cd

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

=>a+c/b+d<c/d (2)

từ 1 và 2 =>đpcm