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

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

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

 Mình chỉ làm được tới khúc này

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:

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

Trường hợp 1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{\left(a+b\right)+\left(a-b\right)}{\left(c+d\right)+\left(c-d\right)}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)

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

Từ (3) và (4) suy ra \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

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

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau

a^2+b^2/c^2+d^2  =   a^2/c^2  =   b^2 / d^2

=>a/c   =    b/d

=>a/b    =    c/d

Chúc bạn học tốt nha

dat k ; ta co a= bk , c=dk , roi tu thay vao ma rut gon nhe

a/b=c/d

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

=>a/b.c/d=(a+b)^2/(c+d)^2

=>ab/cd=(a+b)^2/(c+d)^2  

Vay......

a/b=c/d

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

=> a/b.c/d=(a+b)^2/(c+d)^2

=> ab/cd=(a+b)^2/(c+d)^2

# Hok_tốt nha

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

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

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)

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

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