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

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

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

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

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

+ Nếu \(a+b+c+d\ne0\)

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

\(\Rightarrow c=a\left(đpcm1\right).\)

+ Nếu \(a+b+c+d=0\)

\(\Rightarrow\) hợp với đề.

\(\Rightarrow a+b+c+d=0\left(đpcm2\right).\)

Chúc bạn học tốt!

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

<=> ad + a² + bd + ab = bc + bd + c² + cd

<=> ad + a² + bd + ab - bc - bd - c² - cd = 0

<=> ad + a² + ab - bc - c² - cd = 0

<=> ( ad - cd ) + ( a² - c² ) + ( ab - bc ) = 0

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

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

<=> \(\orbr{\begin{cases}a-c=0\\a+b+c+d=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=c\\a+b+c+d=0\end{cases}\left(đpcm\right)}\)

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

TH1: \(a+b+c+d=0\Rightarrowđpcm\)

TH2: \(a+b+c+d\ne0\Rightarrow\frac{a+b}{b+c}=\frac{c+d}{d+a}=1\)

\(\Rightarrow a+b=b+c\)

\(\Rightarrow a=c\left(đpcm\right)\)

=10000 biết

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

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

=>ad+a^2+bd+ab=bc+bd+c^2+cd

=>ad+a^2+ab=c^2+bc+cd

=>bạn làm tiếp nhé

vì a+b/b+c = c+d/d+a nên

(a+b).(d+a) =(c+d).(b+c)

ad+bd+bd+ab=cb+db+db+dc

ad+ab=cb+dc (  2 vế cùng bớt đi db+db)

a.(d+b)=c.(b+d)

=> a=c

vì a+b/b+c = c+d/d+a nên
(a+b).(d+a) =(c+d).(b+c)
ad+bd+bd+ab=cb+db+db+dc
ad+ab=cb+dc ( 2 vế cùng bớt đi db+db)
a.(d+b)=c.(b+d)
=> a=c

Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)

\(\Leftrightarrow\left(a^2cd-abd^2\right)+\left(b^2cd-abc^2\right)=0\)

\(\Leftrightarrow ad\left(ac-bd\right)-bc\left(ac-bd\right)=0\)

\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (đpcm)

Giải:

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

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

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

Từ (1) và (2) suy ra \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)

Ta có : a/b = c/d suy ra a/c = b/d.

Theo tính chất dãy tỉ số bằng nhau ta có:

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

Suy ra:

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

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

Áp dụng tính chất dãy tỉ số bằng nhau , ta có : 

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

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

Ta có : a/b=c/d<=>a/c=b/d=a+b/c+d=a-b/c-d

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

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

Đặt \(\frac{a}{c}\)=\(\frac{b}{d}\)=k (k\(\in\)Z)\(\Rightarrow\)\(\hept{\begin{cases}a=ck\\b=dk\end{cases}}\) 

\(\Rightarrow\)\(\frac{a+b}{a-b}\)=\(\frac{ck+dk}{ck-dk}\)=\(\frac{k}{k}\).\(\frac{c+d}{c-d}\)=\(\frac{c+d}{c-d}\)

Vậy ta đã chứng minh được \(\frac{a+b}{a-b}\)=\(\frac{c+d}{c-d}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

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

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

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