$\tt{\color{#4B4E53}{୨୧ }\text{ }\color{#4B4E53}{i}\color{#5C5F64}{t}\color{#6E7176}{s}\color{#808388}{Z}\color{#92959A}{u}\color{#A4A7AC}{n}\color{#B6B9BE}{n}}$

$\begin{array}{l} \text{Ta có:} \\ 2x^2 = 5xy + 12y^2 \\ \Leftrightarrow 2x^2 - 5xy - 12y^2 = 0 \\ \Leftrightarrow 2x^2 - 8xy + 3xy - 12y^2 = 0 \\ \Leftrightarrow 2x(x - 4y) + 3y(x - 4y) = 0 \\ \Leftrightarrow (2x + 3y)(x - 4y) = 0 \\ \Leftrightarrow \left[ \begin{array}{l} 2x + 3y = 0 \\ x - 4y = 0 \end{array} \right. \\ \Leftrightarrow \left[ \begin{array}{l} x = -\dfrac{3}{2}y \\ x = 4y \end{array} \right. \\ \\ \text{Trường hợp 1: } x = 4y \\ \text{Thay } x = 4y \text{ vào A, ta được:} \\ A = \dfrac{5(4y)^2 + 4y^2}{4(4y)^2 - y^2} = \dfrac{5 \cdot 16y^2 + 4y^2}{4 \cdot 16y^2 - y^2} = \dfrac{80y^2 + 4y^2}{64y^2 - y^2} = \dfrac{84y^2}{63y^2} = \dfrac{4}{3} \\ \\ \text{Trường hợp 2: } x = -\dfrac{3}{2}y \\ \text{Thay } x = -\dfrac{3}{2}y \text{ vào A, ta được:} \\ A = \dfrac{5 \left(-\dfrac{3}{2}y\right)^2 + 4y^2}{4 \left(-\dfrac{3}{2}y\right)^2 - y^2} = \dfrac{5 \cdot \dfrac{9}{4}y^2 + 4y^2}{4 \cdot \dfrac{9}{4}y^2 - y^2} \\A= \dfrac{\dfrac{45}{4}y^2 + \dfrac{16}{4}y^2}{9y^2 - y^2} = \dfrac{\dfrac{61}{4}y^2}{8y^2} = \dfrac{61}{32} \\ \\ \text{Vậy } A = \dfrac{4}{3} \text{ hoặc } A = \dfrac{61}{32} \end{array}$