Đáp án: A=1/7 hoặc A=-1/7
Giải thích các bước giải:
$\begin{array}{l}
\frac{{{x^2} + {y^2}}}{{xy}} = \frac{{25}}{{12}} \Rightarrow \frac{{{x^2}}}{{xy}} + \frac{{{y^2}}}{{xy}} = \frac{{25}}{{12}}\\
Đặt\Rightarrow \frac{x}{y} + \frac{y}{x} = \frac{{25}}{{12}}\\
:\frac{x}{y} = t \Rightarrow \frac{y}{x} = \frac{1}{t}\\
\Rightarrow t + \frac{1}{t} = \frac{{25}}{{12}}\\
\Rightarrow {t^2} - \frac{{25}}{{12}}t + 1 = 0\\
\Rightarrow 12{t^2} - 25t + 12 = 0\\
\Rightarrow 12{t^2} - 9t - 16t + 12 = 0\\
\Rightarrow 3t\left( {4t - 3} \right) - 4\left( {4t - 3} \right) = 0\\
\Rightarrow \left( {4t - 3} \right)\left( {3t - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
t = \frac{3}{4} = \frac{x}{y} \Rightarrow x = \frac{{3y}}{4} \Rightarrow A = \frac{{\frac{3}{4}y - y}}{{\frac{3}{4}y + y}} = - \frac{1}{7}\\
t = \frac{4}{3} = \frac{x}{y} \Rightarrow x = \frac{{4y}}{3} \Rightarrow A = \frac{{x - y}}{{x + y}} = \frac{{\frac{{4y}}{3} - y}}{{\frac{{4y}}{3} + y}} = \frac{1}{7}
\end{array} \right.
\end{array}$
