\(\left(\sqrt{x+4}\right)\cdot\left(\sqrt{x-4}\right)\)

\(=\sqrt{\left(x+4\right)\cdot\left(x-4\right)}\)

\(=\sqrt{x^2-4x+4x-16}\)

\(=\sqrt{x^2-16}\)

Đặt: a=\(\frac{1}{x}\), b=\(\frac{1}{y}\) ta có:

\(\left\{{}\begin{matrix}a+b=\frac{1}{12}\\4a+6b=\frac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4a+4b=\frac{1}{3}\\4a+6b=\frac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\frac{1}{12}\\-2b=-\frac{1}{15}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{20}\\b=\frac{1}{30}\end{matrix}\right.\)

hay: \(\frac{1}{x}=\frac{1}{20}\Rightarrow x=20\)

\(\frac{1}{y}=\frac{1}{30}\Rightarrow y=30\)