Áp dụng t/c của dãy tỉ số bằng nhau ta có \(\frac{\left(a^{2k}+b^{2k}\right)}{c^{2k}+d^{2k}}=\frac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\frac{\left(a^{2k}+b^{2k}\right)+\left(a^{2k}-b^{2k}\right)}{\left(c^{2k}+d^{2k}\right)+\left(c^{2k}-d^{2k}\right)}=\frac{\left(a^{2k}+b^{2k}\right)-\left(a^{2k}-b^{2k}\right)}{\left(c^{2k}+d^{2k}\right)-\left(c^{2k}-d^{2k}\right)}\)

=> \(\frac{a^{2k}}{c^{2k}}=\frac{b^{2k}}{d^{2k}}\) => \(\left(\frac{a}{c}\right)^{2k}=\left(\frac{b}{d}\right)^{2k}\) => \(\frac{a}{c}=\frac{b}{d}\) hoặc \(\frac{a}{c}=-\frac{b}{d}\) ( do số mũ 2k chẵn)

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

ĐKXĐ: \(b,d\ne0,c\ne\pm d\)

Áp dụng t/c dtsbn:

\(\dfrac{a^{2k}+b^{2k}}{c^{2k}+d^{2k}}=\dfrac{a^{2k}-b^{2k}}{c^{2k}-d^{2k}}=\dfrac{a^{2k}+b^{2k}+a^{2k}-b^{2k}}{c^{2k}+d^{2k}+c^{2k}-d^{2k}}=\dfrac{2a^{2k}}{2c^{2k}}=\dfrac{a^{2k}}{c^{2k}}\left(1\right)\)

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

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^{2k}}{c^{2k}}=\dfrac{b^{2k}}{d^{2k}}\Rightarrow\dfrac{a^{2k}}{b^{2k}}=\dfrac{c^{2k}}{d^{2k}}\Rightarrow\dfrac{a}{b}=\pm\dfrac{c}{d}\left(đpcm\right)\)

Cảm ơn bạn

theo mik là vì dạng 2 là TBC của số chẵn nên fai là 2k