Từ giả thiết: \(\frac{a}{b}=\frac{c}{d}\)=>ad=bc                                                  (1)

Ta có: ab(c²-d²)=abc²-abd²=acbc-adbd                                             (2)

          cd(a²-b²)=a²cd-b²cd=acad-bcbd                                             (3)

Từ (1) ,(2),(3)=> ab(c²-d²)=cd(a²-b²)=>\(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)            (đpcm)

Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

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

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

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

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

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

Chúc bạn học tốt!

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}< =>\left(a^2+b^2\right).cd=\left(c^2+d^2\right).ab< =>\left(a^2+b^2\right).cd-\left(c^2+d^2\right).ab=0\)

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

\(< =>ac\left(ad-bc\right)-bd\left(ad-bc\right)=0< =>\left(ad-bc\right)\left(ac-bd\right)=0< =>ad-bc=0< =>ad=bc< =>\frac{a}{d}=\frac{b}{c}\) (đpcm)

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

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)(Tính chất dãy tỉ số bằng nhau)

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

\(\frac{a}{b}=\frac{c}{d}\) => \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)(Tính chất dãy tỉ số bằng nhau)

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

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

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

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

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

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

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

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

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