Question

cho \(\frac{a}{b}=\frac{c}{d}\left(b,d\ne0;b\ne d,-d\right)\)

Chứng minh \(\left(\frac{a-b}{c-d}\right)^{2018}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)

 

Answer 1

với c=0=>a=0 đẳng thức đúng

với c khác 0 ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{\left(a-b\right)^{2018}}{\left(c-d\right)^{2018}}=\frac{a^{2018}}{c^{2018}}=\frac{b^{2018}}{d^{2018}}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)

=>\(\frac{\left(a-b\right)^{2018}}{\left(c-d\right)^{2018}}=\frac{a^{2018}+b^{2018}}{c^{2018}+d^{2018}}\)