Giải thích các bước giải:
Có:
\(\begin{array}{l}
y' = \frac{{2\left( {3x - 1} \right) - 3\left( {2x + 1} \right)}}{{{{\left( {3x - 1} \right)}^2}}}\\
= \frac{{6x - 2 - 6x - 3}}{{{{\left( {3x - 1} \right)}^2}}} = \frac{{ - 5}}{{{{\left( {3x - 1} \right)}^2}}}
\end{array}\)
Gọi tiếp điểm là \(M\left( {{x_0};{y_0}} \right)\)
\(\begin{array}{l}
Do:k = y'\left( {{x_0}} \right) = - 1\\
\to \frac{{ - 5}}{{{{\left( {3{x_0} - 1} \right)}^2}}} = - 1\\
\to 5 = 9{x_0}^2 - 6{x_0} + 1\\
\to \left[ \begin{array}{l}
{x_0} = \frac{{1 + \sqrt 5 }}{3}\\
{x_0} = \frac{{1 - \sqrt 5 }}{3}
\end{array} \right. \to \left[ \begin{array}{l}
{y_0} = \frac{{2 + \sqrt 5 }}{3}\\
{y_0} = \frac{{2 - \sqrt 5 }}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - 1.\left( {x - \frac{{1 + \sqrt 5 }}{3}} \right) + \frac{{2 + \sqrt 5 }}{3}\\
y = - 1\left( {x - \frac{{1 - \sqrt 5 }}{3}} \right) + \frac{{2 - \sqrt 5 }}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = - x + \frac{{3 + 2\sqrt 5 }}{3}\\
y = - x + \frac{{3 - 2\sqrt 5 }}{3}
\end{array} \right.
\end{array}\)
