Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(a=bk;c=dk\)
Suy ra :
\(\frac{a^{2019}+12b^{2019}}{c^{2019}+12d^{2019}}=\frac{\left(bk\right)^{2019}+12b^{2019}}{\left(dk\right)^{2019}+12d^{2019}}=\frac{b^{2019}.k^{2019}+12b^{2019}}{d^{2019}.k^{2019}+12d^{2019}}=\frac{b^{2019}\left(k^{2019}+12\right)}{d^{2019}\left(k^{2019}+12\right)}\)
\(\frac{b^{2019}}{k^{2019}}\left(1\right)\)
\(\text{⋆}\frac{\left(12a-11b\right)^{2019}}{\left(12c-11d\right)^{2019}}=\frac{\left(12bk-11b\right)^{2019}}{\left(12dk-11d\right)^{2019}}=\frac{\left[b\left(12k-11b\right)\right]^{2019}}{\left[b\left(12k-11d\right)\right]}=\frac{b^{2019}.\left(12k-11\right)^{2019}}{d^{2019}.\left(12k-11\right)^{2019}}\)
\(=\frac{b^{2019}}{d^{2019}}\)
Từ (1) và (2) suy ra : \(\frac{a^{2019}+12b^{2019}}{c^{2019}+12d^{2019}}=\frac{\left(12a-11b\right)^{2019}}{\left(12c-11d\right)^{2019}}\left(đpcm\right)\)
