Đáp án:
\(\left[ \begin{array}{l}
y = 0,3056448602\\
y = - 0,3115972412
\end{array} \right. \to \left[ \begin{array}{l}
x = - 0,5279563871\\
x = 0,7065278157
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{{16}}\\
3x + 6y = \dfrac{1}{4}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \left( {\dfrac{1}{4} - 6y} \right):3\\
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{{16}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{1 - 24y}}{{12}}\\
\dfrac{1}{{\dfrac{{1 - 24y}}{{12}}}} + \dfrac{1}{y} = \dfrac{1}{{16}}\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to \dfrac{{12}}{{1 - 24y}} + \dfrac{1}{y} = \dfrac{1}{{16}}\\
\to \dfrac{{12y + 1 - 24y}}{{y\left( {1 - 24y} \right)}} = \dfrac{1}{{16}}\\
\to 16\left( {1 - 12y} \right) = y - 24{y^2}\\
\to 16 - 192{y^2} = y - 24{y^2}\\
\to 168{y^2} + y - 16 = 0\\
\to \left[ \begin{array}{l}
y = 0,3056448602\\
y = - 0,3115972412
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 0,5279563871\\
x = 0,7065278157
\end{array} \right.
\end{array}\)
