[กขฃคฅฆง]

15. จากโจทย์ได้

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \rightarrow a_1=ka_2 , b_1=kb_2$

$\frac{a_2}{a_3} = \frac{b_2}{b_3} $

$\dfrac{(a_1^2+b_1^2)(a_2^2+3b_2^2)}{(a_1^2+3b_1^2)(a_2^2+b_2^2)} = \dfrac{k^2(a_2^2+b_2^2)(a_2^2+3b_2^2)}{k^2(a_2^2+3b_2^2)(a_2^2+b_2^2)} = 1 =(\dfrac{a_1b_2}{a_2b_1})(\dfrac{b_2^2a_3^2}{a_2^2b_3^2}) = \dfrac{a_1b_2^3a_3^2}{b_1a_2^3b_3^2}$