Đáp án+Giải thích các bước giải:
Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.$
Ta có:
$\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}$
$\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}$
$\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}$ (đpcm)
