Sửa lại đề: \(\dfrac{a}{b}=\dfrac{c}{d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(VT=\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{ck+dk}{c+d}\right)^2=\left(\dfrac{k.\left(c+d\right)}{c+d}\right)^2=k^2\left(1\right)\)
\(VP=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(ck\right)^2+\left(dk\right)^2}{c^2+d^2}=\dfrac{c^2.k^2+d^2.k^2}{c^2+d^2}=\dfrac{k^2.\left(c^2+d^2\right)}{c^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\left(đpcm\right)\)
Vậy \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
