Ta có :

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\left(\frac{a+b}{c+d}\right)^2\left(1\right)\)

\(\frac{a^2+b^2}{c^2+d^2}-\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)

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

TH1 : \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{b}\left(3\right)\)

        \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)

từ ( 3 ) và ( 4 ) suy ra : \(\frac{a}{c}=\frac{b}{d}\text{ hay }\frac{a}{b}=\frac{c}{d}\)

TH2 : \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)+\left(b-a\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2b}{2c}=\frac{b}{c}\left(5\right)\)

     \(\frac{a+b}{c+d}=\frac{b-a}{c-d}=\frac{\left(a+b\right)-\left(b-a\right)}{\left(c+d\right)-\left(c-d\right)}=\frac{2a}{2d}=\frac{a}{d}\left(6\right)\)

Từ ( 5 ) và ( 6 ) suy ra : \(\frac{b}{c}=\frac{a}{d}\text{ hay }\frac{a}{b}=\frac{d}{c}\)

Vậy : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\text{ thì }\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)

kinh quá

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

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

\(ad+a^2+bd+ab=bc+bd+c^2+cd\)

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

\(a+a^2=c+c^2\)

\(a=c\)

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

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

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

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

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

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

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

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

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

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

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

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

=> \(a^2+ab+ad+bd=c^2+bc+bd+cd\)

=>\(a^2+ab+ad-bc-c^2-cd=0\)

=>\(\left(a^2-c^2\right)+\left(ab-cd\right)+\left(ab-ac\right)=0\)

=>\(\left(a-c\right)\left(a+c\right)+d\left(a-c\right)+b\left(a-c\right)=0\)

=>\(\left(a-c\right)\left(a+b+c+d\right)=0\)

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

hacker 2k6

bổ sung chỗ dưới

bạn => thêm là

\(=>\orbr{\begin{cases}a=c\\a+b+c+d=0\end{cases}\left(dpcm\right)}\)

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

<=> cd(a² + b²) = ab(c² + d²) 

<=> a²cd  + b²cd - abc² - abd² = 0

<=> (a²cd - abc²) + (b²cd - abd²) = 0

<=> ac(ad - bc) + bd(bc - ad) = 0 

<=> ac(ad - bc) - bd(ad - bc) = 0

<=> (ac - bd)(ad - bc) = 0

<=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{d}=\frac{b}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\left(\text{đpcm}\right)\)