Đáp án:
\(x \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0\\
\dfrac{{1 + 2x}}{{36}} = \dfrac{{1 + 4x}}{{48}} = \dfrac{{1 + 6x}}{{6x}}\\
\to \left\{ \begin{array}{l}
\dfrac{{1 + 2x}}{{36}} = \dfrac{{1 + 4x}}{{48}}\\
\dfrac{{1 + 2x}}{{36}} = \dfrac{{1 + 6x}}{{6x}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{8\left( {1 + 2x} \right)}}{{36.8}} = \dfrac{{6\left( {1 + 4x} \right)}}{{48.6}}\\
\dfrac{{x\left( {1 + 2x} \right)}}{{36x}} = \dfrac{{6\left( {1 + 6x} \right)}}{{36x}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{8 + 16x}}{{288}} = \dfrac{{6 + 24x}}{{288}}\\
x + 2{x^2} = 6 + 36x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8x = 2\\
2{x^2} - 35x - 6 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{4}\\
\left[ \begin{array}{l}
x = 17,66978139\\
x = - 0,1697813875
\end{array} \right.
\end{array} \right.\left( {vô lý} \right)\\
\to x \in \emptyset
\end{array}\)
