Đáp án:
$\left( {x;y} \right) \in \left\{ {\left( {1;\dfrac{{ - 1}}{7}} \right);\left( {\dfrac{{ - 6}}{7};\dfrac{{ - 1}}{7}} \right)} \right\}$
Giải thích các bước giải:
ĐKXĐ:$x \ne 0$
Ta có;
$\dfrac{{x + y}}{3} = \dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 5y}}{x}\left( 1 \right)$
Từ $(1)$ ta có:
$\begin{array}{l}
\dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 5y}}{x}\\
\Leftrightarrow \dfrac{{1 + 3y}}{{2x}} = \dfrac{{2 + 10y}}{{2x}}\\
\Leftrightarrow \dfrac{{2 + 10y}}{{2x}} - \dfrac{{1 + 3y}}{{2x}} = 0\\
\Leftrightarrow \dfrac{{1 + 7y}}{{2x}} = 0\\
\Leftrightarrow 1 + 7y = 0\\
\Leftrightarrow y = \dfrac{{ - 1}}{7}
\end{array}$
Khi đó thay vào $(1)$ ta có: $\dfrac{{1 + 3y}}{{2x}} = \dfrac{{1 + 3.\left( {\dfrac{{ - 1}}{7}} \right)}}{{2x}} = \dfrac{2}{{7x}}$
Và:
$\begin{array}{l}
\dfrac{{x + \left( {\dfrac{{ - 1}}{7}} \right)}}{3} = \dfrac{2}{{7x}}\\
\Leftrightarrow \dfrac{{7x - 1}}{{21}} = \dfrac{2}{{7x}}\\
\Leftrightarrow \dfrac{{7x - 1}}{{21}} - \dfrac{2}{{7x}} = 0\\
\Leftrightarrow \dfrac{{x\left( {7x - 1} \right) - 2.3}}{{21x}} = 0\\
\Leftrightarrow \dfrac{{7{x^2} - x - 6}}{{21x}} = 0\\
\Rightarrow 7{x^2} - x - 6 = 0\\
\Leftrightarrow \left( {7{x^2} - 7x} \right) + \left( {6x - 6} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {7x + 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{ - 6}}{7}
\end{array} \right.\left( {tm} \right)
\end{array}$
Vậy $\left( {x;y} \right) \in \left\{ {\left( {1;\dfrac{{ - 1}}{7}} \right);\left( {\dfrac{{ - 6}}{7};\dfrac{{ - 1}}{7}} \right)} \right\}$ thỏa mãn
