Giải thích các bước giải:
a,
Ta có:
\[\begin{array}{l}
\frac{x}{2} = \frac{y}{5} \Rightarrow \frac{x}{2}.\frac{x}{2} = \frac{y}{5}.\frac{x}{2}\\
\Leftrightarrow {\left( {\frac{x}{2}} \right)^2} = \frac{{xy}}{{10}}\\
\Leftrightarrow {\left( {\frac{x}{2}} \right)^2} = \frac{{90}}{{10}}\\
\Leftrightarrow {\left( {\frac{x}{2}} \right)^2} = 9\\
\Rightarrow \left[ \begin{array}{l}
\frac{x}{2} = 3\\
\frac{x}{2} = - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 6 \Rightarrow y = 15\\
x = - 6 \Rightarrow y = - 15
\end{array} \right.
\end{array}\]
b,
\[\begin{array}{l}
{\left( {\frac{1}{3} - 2x} \right)^{102}} + {\left( {3y - x} \right)^{104}} = 0\\
\left\{ \begin{array}{l}
{\left( {\frac{1}{3} - 2x} \right)^{102}} \ge 0\\
{\left( {3y - x} \right)^{104}} \ge 0
\end{array} \right. \Rightarrow {\left( {\frac{1}{3} - 2x} \right)^{102}} + {\left( {3y - x} \right)^{104}} \ge 0\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{3} - 2x = 0\\
3y - x = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{1}{6}\\
y = \frac{1}{{18}}
\end{array} \right.
\end{array}\]
