Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\frac{x}{5} = \frac{y}{9}\\
\Rightarrow {\left( {\frac{x}{5}} \right)^2} = \frac{x}{5}.\frac{x}{5} = \frac{x}{5}.\frac{y}{9} = \frac{{xy}}{{45}} = \frac{{405}}{{45}} = 9\\
\Leftrightarrow \left[ \begin{array}{l}
\frac{x}{5} = \frac{y}{9} = 3\\
\frac{x}{5} = \frac{y}{9} = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 15\\
y = 27
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 15\\
y = - 27
\end{array} \right.
\end{array} \right.\\
b,\\
\frac{{1 + 5y}}{{24}} = \frac{{1 + 7y}}{{7x}} = \frac{{1 + 9y}}{{2x}}\\
\frac{{1 + 7y}}{{7x}} = \frac{{1 + 9y}}{{2x}}\\
\Leftrightarrow \frac{{2.\left( {1 + 7y} \right)}}{{2.7x}} = \frac{{7.\left( {1 + 9y} \right)}}{{7.2x}}\\
\Leftrightarrow \frac{{2 + 14y}}{{14}} = \frac{{7 + 63y}}{{14x}}\\
\Leftrightarrow 2 + 14y = 7 + 63y\\
\Leftrightarrow y = - \frac{5}{{49}}\\
\frac{{1 + 5y}}{{24}} = \frac{{1 + 7y}}{{7x}} \Leftrightarrow \frac{{1 + 5.\frac{{ - 5}}{{49}}}}{{24}} = \frac{{1 + 7.\frac{{ - 5}}{{49}}}}{{7x}} \Rightarrow x = 2
\end{array}\)
