Ta có : a+b/b+c = c+d/d+a 
=> (a+b)/(c+d)= (b+c)/(d+a) 
=> (a+b)/(c+d)+1=(b+c)/(d+a)+1 
hay: (a+b+c+d)/(c+d)=(b+c+d+a)/(d+a) 
- Nếu a+b+c+d khác 0 thì : c+d=d+a => c=a (1)
- Nếu a+b+c+d = 0 (2) 

Từ (1) và (2) 

\(\RightarrowĐPCM\)

Ta có : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

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

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

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

Nếu a + b + c + d khác 0 thì c + d = d + a => c = a ( hoặc a = c )

Nếu a + b + c + d = 0 ( đpcm )

Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

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

\(\implies\) \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)

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

\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)

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

Có:

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

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

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

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

\(\left(1\right)\left(2\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

Cách 1: 

Ta có:

Cách 2:

Từ 

hay 

ta có:a/b=c/d

=>a/c=b/d

áp dụng tích chất dãy tỉ số bằng nhau ta có:

a/c=b/d=a+b/c+d=a-c/c-d

=>a+b/c+d=a-b/c-d

do đó: a+b/a-c=c+d/c-d

ta có;

a/b=c/d =>a/c=b/d

áp dụng tích chất dãy tỉ số bằng nhau ta có:

a/c=b/d =>a+b/c+d=a-b/c-d

=>a+b/c+d=a-b/c-d  => a+b/a-b=c+d/c-d

đập chết cha mày bây giờ đưa tiền đây:500triệu mày nợ mua đất 

chỉ cần thừa nhận không cần chứng minh

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

\(\Rightarrow\hept{\begin{cases}c+d=\left(d+a\right)k\\a+b=\left(b+c\right)k\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}c+d=dk+ak\\a+b=bk+ck\end{cases}}\)

\(\Rightarrow a+b+c+d=bk+ck+dk+ak\)

\(\Rightarrow a+b+c+d=\left(a+b+c+d\right)k\)

\(\Rightarrow k=1\)

\(\Rightarrow\hept{\begin{cases}c+d=d+a\\a+b=b+c\end{cases}}\)

\(\Rightarrow c+d-d-a=0\)

\(\Rightarrow c-a=0\)

\(\Rightarrow c=a\)

\(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)

\(\left(a+d\right)^2-\left(a-d\right)^2=\left(b+c\right)^2-\left(b-c\right)^2\)

\(\left(a+d-a+d\right)\left(a+d+a-d\right)=\left(b+c-b+c\right)\left(b+c+b-c\right)\)

\(2d\times2a=2b\times2c\)

\(ad=bc\)

\(\frac{a}{c}=\frac{b}{d}\left(\text{đ}pcm\right)\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

Ta có:\(\frac{a+b}{a-b}=\frac{bk+b}{bk-b}=\frac{b\left(k+1\right)}{b\left(k-1\right)}=\frac{k+1}{k-1}\left(1\right)\)

\(\frac{c+d}{c-d}=\frac{dk+d}{dk-d}=\frac{d\left(k+1\right)}{d\left(k-1\right)}=\frac{k+1}{k-1}\left(2\right)\)

Từ (1) và (2), ta có: \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

_Học tốt_

Thanks