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

Ta có  \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\)(dãy tỉ số bằng nhau)

=> \(\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b-c}{b+c-d}\right)^3\)

=> \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)

=> \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)(dãy tỉ số bằng nhau)

=> \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)(đpcm)

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

Sử dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{a+b+c-\left(a-b+c\right)}{a+b-c-\left(a-b-c\right)}=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)

=>\(\frac{a+b+c}{a+b-c}=1\)

=>a+b+c=a+b-c

=>c+c=a+b-a-b

=>2c=0

=>c=0

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:

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