Question

giup to voi mn 6h30 phải nộp:((

Answer 1

mn ơi:(

Answer 2

Bài 57:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{k^2b^2+k^2d^2}{b^2+d^2}=k^2\)

\(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)

Do đó: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)