Question

Chứng minh từ \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\);b+d \(\ne\)0

ta có thể suy ra \(\dfrac{ac}{bd}\)=\(\dfrac{5a^2+7c^2}{5b^2+7d^2}\)

Answer 1

Giải:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\) (1)

\(\dfrac{5a^2+7c^2}{5b^2+7d^2}=\dfrac{5\left(bk\right)^2+7\left(dk\right)^2}{5b^2+7d^2}=\dfrac{5b^2.k^2+7d^2.k^2}{5b^2+7d^2}=\dfrac{k^2\left(5b^2+7d^2\right)}{5b^2+7d^2}=k^2\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{ac}{bd}=\dfrac{5a^2+7c^2}{5b^2+7d^2}\left(đpcm\right)\)