Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0
\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}
a = \frac{1}{x}\\
b = \frac{1}{y}
\end{array} \right.\), khi đó hệ pt đã cho trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
6a - 4b = - 4\\
3a + 8b = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3a - 2b = - 2\\
3a + 8b = 3
\end{array} \right.\\
\Rightarrow \left( {3a + 8b} \right) - \left( {3a - 2b} \right) = 3 - \left( { - 2} \right)\\
\Leftrightarrow 10b = 5\\
\Leftrightarrow b = \frac{1}{2} \Rightarrow a = - \frac{1}{3}\\
\Rightarrow \left\{ \begin{array}{l}
\frac{1}{x} = - \frac{1}{3}\\
\frac{1}{y} = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 3\\
y = 2
\end{array} \right.
\end{array}\)
c,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
y \ne 0
\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}
a = \frac{1}{x}\\
b = \frac{1}{y}
\end{array} \right.\), khi đó hệ pt đã cho trở thành:
\(\begin{array}{l}
\left\{ \begin{array}{l}
2a + 3b = - \frac{1}{2}\\
- a + 6b = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2a + 3b = - \frac{1}{2}\\
a = 6b - \frac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2\left( {6b - \frac{1}{2}} \right) + 3b = - \frac{1}{2}\\
a = 6b - \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
b = \frac{1}{{30}}\\
a = - \frac{3}{{10}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\frac{1}{x} = - \frac{3}{{10}}\\
\frac{1}{y} = \frac{1}{{30}}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - \frac{{10}}{3}\\
y = 30
\end{array} \right.
\end{array}\)
