Đáp án:
\[A = \pm \frac{1}{7}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{{x^2} + {y^2}}}{{xy}} = \frac{{25}}{{12}}\\
\Leftrightarrow 12{x^2} + 12{y^2} = 25xy\\
\Leftrightarrow \left( {12{x^2} - 16xy} \right) - \left( {9xy - 12{y^2}} \right) = 0\\
\Leftrightarrow 4x\left( {3x - 4y} \right) - 3y\left( {3x - 4y} \right) = 0\\
\Leftrightarrow \left( {4x - 3y} \right)\left( {3x - 4y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x - 3y = 0\\
3x - 4y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{3}{4}y\\
x = \frac{4}{3}y
\end{array} \right.\\
TH1:\,\,\,x = \frac{3}{4}y\\
\Rightarrow A = \frac{{x - y}}{{x + y}} = \frac{{\frac{3}{4}y - y}}{{\frac{3}{4}y + y}} = - \frac{{\frac{1}{4}y}}{{\frac{7}{4}y}} = \frac{{ - 1}}{7}\\
TH2:\,\,\,\,\,x = \frac{4}{3}y\\
\Rightarrow A = \frac{{\frac{4}{3}y - y}}{{\frac{4}{3}y + y}} = \frac{{\frac{1}{3}y}}{{\frac{7}{3}y}} = \frac{1}{7}\\
\Rightarrow A = \pm \frac{1}{7}
\end{array}\)
