Đáp án:
a. \(\left\{ \begin{array}{l}
y = - 12\\
x = 4
\end{array} \right.\)
b. \(\left\{ \begin{array}{l}
x = 2\\
y = 2
\end{array} \right.\)
c. \(\left\{ \begin{array}{l}
x = 0\\
y = - \frac{3}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\frac{1}{x} + \frac{1}{y} = \frac{1}{6}\\
\frac{3}{x} + \frac{6}{y} = \frac{1}{4}
\end{array} \right. \to \left\{ \begin{array}{l}
\frac{{ - 3}}{x} - \frac{3}{y} = - \frac{3}{6}\\
\frac{3}{x} + \frac{6}{y} = \frac{1}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\frac{3}{y} = - \frac{1}{4}\\
\frac{1}{x} + \frac{1}{y} = \frac{1}{6}
\end{array} \right. \to \left\{ \begin{array}{l}
y = - 12\\
x = 4
\end{array} \right.\left( {TM} \right)\\
b.DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\frac{1}{x} + \frac{1}{y} = 1\\
\frac{4}{x} - \frac{2}{y} = 1
\end{array} \right. \to \left\{ \begin{array}{l}
\frac{2}{x} + \frac{2}{y} = 2\\
\frac{4}{x} - \frac{2}{y} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\frac{6}{x} = 3\\
\frac{4}{x} - \frac{2}{y} = 1
\end{array} \right. \to \left\{ \begin{array}{l}
x = 2\\
y = 2
\end{array} \right.\left( {TM} \right)\\
c.DK:x \ne - 1;y \ne - 1\\
\left\{ \begin{array}{l}
\frac{2}{{x + 1}} - \frac{1}{{y + 1}} = 4\\
\frac{1}{{x + 1}} + \frac{1}{{y + 1}} = - 1
\end{array} \right. \to \left\{ \begin{array}{l}
\frac{3}{{x + 1}} = 3\\
\frac{2}{{x + 1}} - \frac{1}{{y + 1}} = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + 1 = 1\\
\frac{2}{{x + 1}} - \frac{1}{{y + 1}} = 4
\end{array} \right. \to \left\{ \begin{array}{l}
x = 0\\
y = - \frac{3}{2}
\end{array} \right.\left( {TM} \right)
\end{array}\)
