Bạn tham khảo!
$Từ$ $ad$ $=$ $bc$
$Đặt$ $\dfrac{a}{b}$ $=$ $\dfrac{c}{d}$ $=$ $k$
$⇒$ $\left \{ {{\dfrac{a}{b}=k⇒a=bl} \atop {\dfrac{c}{d}= k ⇒ c = dk}} \right.$
$Thay$ $vào$ $ta$ $có$ $:$
$\dfrac{3a^2-5b^2}{3c^2-5d^2}$ $=$ $\dfrac{3c^2·k^2-5d^2·k^2}{3c^2-5d^2}$ $=$ $\dfrac{k^2·(3c^2-5d^2)}{3c^2-5d^2}$ $=$ $k^2$ $(1)$
$Thay$ $vào$ $ta$ $có$ $:$
$(\dfrac{a+b}{c+d})^2$ $=$ $(\dfrac{ck+dk}{c+a})^2$ $=$ $\dfrac{c^2·k^2+d^2·k^2}{c^2+d^2}$ $=$ $\dfrac{k^2·(3c^2+d^2)}{c^2+d^2}$ $=$ $k^2$ $(2)$
$Từ$ $(1)$ $và$ $(2)$ $⇒$ $\dfrac{3a^2-5b^2}{3c^2-5d^2}$ = $(\dfrac{a+b}{c+d})^2$
Xin hay nhất!
$@$ $King$ $Of$ $Rap$
