Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có: \(\left(\frac{a-b}{c-d}\right)^4=\left(\frac{bk-b}{dk-d}\right)^4=\left[\frac{b\left(k-1\right)}{d\left(k-1\right)}\right]^4=\left(\frac{b}{d}\right)^4\) (1)
\(\frac{a^4+b^4}{c^4+d^4}=\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4.k^4+b^4}{d^4.k^4+d^4}=\frac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\frac{b^4}{d^4}=\left(\frac{b}{d}\right)^4\) (2)
Từ (1) và (2) \(\Rightarrow\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\left(đpcm\right)\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^4}{c^4}=\frac{b^4}{d^4}=\left(\frac{a-b}{c-d}\right)^4\) (1)
Ta lại có:
\(\frac{a^4}{c^4}=\frac{b^4}{d^4}=\frac{a^4+b^4}{c^4+d^4}\) (2)
Từ (1);(2)\(\Rightarrow\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)