Giải thích các bước giải:
b,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0
\end{array} \right.\)
Đặt \(a = \frac{1}{x};\,\,\,b = \frac{1}{y}\) thì hệ pt đã cho trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
5a - 3b = 0\\
2a - b = \frac{1}{{30}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
5a - 3b = 0\\
b = 2a - \frac{1}{{30}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
5a - 3\left( {2a - \frac{1}{{30}}} \right) = 0\\
b = 2a - \frac{1}{{30}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \frac{1}{{10}}\\
b = \frac{1}{6}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\frac{1}{x} = \frac{1}{{10}}\\
\frac{1}{y} = \frac{1}{6}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 10\\
y = 6
\end{array} \right.
\end{array}\)
d,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 2\\
y \ne - 2
\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}
a = \frac{1}{{x - 2}}\\
b = \frac{1}{{y + 2}}
\end{array} \right.\) khi đó, hệ pt đã cho trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
4a - 5b = \frac{5}{2}\\
3a + b = \frac{7}{5}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4a - 5b = \frac{5}{2}\\
b = \frac{7}{5} - 3a
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4a - 5.\left( {\frac{7}{5} - 3a} \right) = \frac{5}{2}\\
b = \frac{7}{5} - 3a
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \frac{1}{2}\\
b = - \frac{1}{{10}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{{x - 2}} = \frac{1}{2}\\
\frac{1}{{y + 2}} = - \frac{1}{{10}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = - 12
\end{array} \right.
\end{array}\)
