Lời giải:ĐK: \(x \neq 0 ; y \neq 0\)
Đặt \(\frac{1}{x}=a ; \frac{1}{y}=b\)
Khi đó ta có:
\(\left\{\begin{array}{l} \frac{1}{x}+\frac{1}{y}=\frac{1}{3} \\ \frac{1}{x}-\frac{1}{y}=\frac{1}{12} \end{array} \Leftrightarrow\left\{\begin{array}{l} a+b=\frac{1}{3} \\ a-b=\frac{1}{12} \end{array} \Leftrightarrow\left\{\begin{array}{l} \left(b+\frac{1}{12}\right)+b=\frac{1}{3} \\ a=b+\frac{1}{12} \end{array}\right.\right.\right.\)
\(\Leftrightarrow\left\{\begin{array}{l} 2 b=\frac{1}{3}-\frac{1}{12} \\ a=b+\frac{1}{12} \end{array} \Leftrightarrow\left\{\begin{array}{c} 2 b=\frac{1}{4} \\ a=b+\frac{1}{12} \end{array} \Leftrightarrow\left\{\begin{array}{l} b=\frac{1}{8} \\ a=\frac{1}{8}+\frac{1}{12} \end{array} \Leftrightarrow\left\{\begin{array}{l} b=\frac{1}{8} \\ a=\frac{5}{24} \end{array}\right.\right.\right.\right.\)
Với \(\left\{\begin{array}{l} b=\frac{1}{8} \\ a=\frac{5}{24} \end{array} \Leftrightarrow\left\{\begin{array}{l} \frac{1}{y}=\frac{1}{8} \\ \frac{1}{x}=\frac{5}{24} \end{array} \Leftrightarrow\left\{\begin{array}{l} x=\frac{24}{5} \\ y=8 \end{array}\right.\right.\right.\)
Vậy, nghiệm của hệ phương trình là: \(\left(\frac{24}{5} ; 8\right)\)
