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