Đáp án:
\[\left[ \begin{array}{l}
A = \frac{2}{5}\\
A = \frac{5}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2} + xy - 12{y^2} = 0\\
\Leftrightarrow \left( {{x^2} + 4xy} \right) - \left( {3xy + 12{y^2}} \right) = 0\\
\Leftrightarrow x\left( {x + 4y} \right) - 3y\left( {x + 4y} \right) = 0\\
\Leftrightarrow \left( {x + 4y} \right)\left( {x - 3y} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 4y = 0\\
x - 3y = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 4y\\
x = 3y
\end{array} \right.
\end{array}\)
Nếu \(x = - 4y\) ta có:
\(A = \frac{{x + 2y}}{{x - y}} = \frac{{ - 4y + 2y}}{{ - 4y - y}} = \frac{{ - 2y}}{{ - 5y}} = \frac{2}{5}\)
Nếu \(x = 3y\) ta có:
\(A = \frac{{x + 2y}}{{x - y}} = \frac{{3y + 2y}}{{3y - y}} = \frac{{5y}}{{2y}} = \frac{5}{2}\)
Vậy \(\left[ \begin{array}{l}
A = \frac{2}{5}\\
A = \frac{5}{2}
\end{array} \right.\)
