ĐK:\(x,y\ne0\). Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\).
Ta có hệ phương trình: \(\left\{ \begin{array}{l} a + b = \dfrac{2}{4}\\ \dfrac{1}{6}.a + \dfrac{1}{5}.b = \dfrac{2}{{15}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \dfrac{1}{2}\\ b = \dfrac{1}{4} \end{array} \right.\)
\(\Rightarrow \left\{ \begin{array}{l} \dfrac{1}{x} = \dfrac{1}{2}\\ \dfrac{1}{y} = \dfrac{1}{4} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = 4 \end{array} \right.\left( {tm} \right)\)
ĐKXĐ: \(x,y\ne0\)
Đặt \(\frac{1}{x}=a,\frac{1}{y}=b\)
Hệ trở thành: \(\left\{{}\begin{matrix}a+b=\frac{3}{4}\\\frac{a}{6}+\frac{b}{5}=\frac{2}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+4b=3\\5a+6b=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}20a+20b=15\\20a+24b=16\end{matrix}\right.\Rightarrow4b=1\Leftrightarrow b=\frac{1}{4}\);\(a=\frac{1}{2}\)
suy ra \(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
