Question

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

\(\frac{\left(a+b\right)^{2019}}{\left(c+d\right)^{2019}}=\frac{a^{2019}+c^{2019}}{b^{2019}+d^{2019}}\)

 

Answer 1

Sửa đề chút:

-Cho tỉ lệ thức

-Yêu cầu CM tỉ lệ thức kia

Answer 2

Đặt  \(\frac{a}{b}=\frac{c}{d}=k\)

 \(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\frac{a^{2019}+c^{2019}}{b^{2019}+d^{2019}}=\frac{\left(bk\right)^{2019}+\left(dk\right)^{2019}}{b^{2019}+d^{2019}}=\frac{b^{2019}.k^{2019}+d^{2019}.k^{2019}}{b^{2019}+d^{2019}}=\frac{k^{2019}.\left(b^{2019}+d^{2019}\right)}{b^{2019}+d^{2019}}=k^{2019}\)(1)

\(\frac{\left(a+c\right)^{2019}}{\left(b+d\right)^{2019}}=\frac{\left(bk+dk\right)^{2019}}{\left(b+d\right)^{2019}}=\frac{[k.\left(b+d\right)]^{2019}}{\left(b+d\right)^{2019}}=\frac{k^{2019}.\left(b+d\right)^{2019}}{\left(b+d\right)^{2019}}=k^{2019}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{a^{2019}+c^{2019}}{b^{2019}+d^{2019}}=\frac{\left(a+c\right)^{2019}}{\left(b+d\right)^{2019}}\)

Mình viết sai đề đó nha