Question

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

Answer 1

Thiếu đề, mk sửa lại

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

Đặt :

\(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

Ta có :

\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(ck\right)^2+c^2}{\left(dk\right)^2+d^2}=\dfrac{c^2.k^2+c^2}{d^2.k^2+d^2}=\dfrac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{c^2}{d^2}\)

\(VT=\dfrac{ac}{bd}=\dfrac{ck.c}{dk.d}=\dfrac{c^2}{d^2}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)