Đáp án:
b. \(\left[ \begin{array}{l}
y = \frac{7}{3}x - \frac{{218}}{{27}}\\
y = \frac{7}{3}x + \frac{{38}}{{27}}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
y' = 3{x^2} - 6x\\
\to y'\left( {{x_0}} \right) = 3{x_0}^2 - 6{x_0}\\
a.Do:{x_0} = - 1\\
\to {y_0} = {\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} + 1 = - 3\\
y'\left( { - 1} \right) = k = 3{\left( { - 1} \right)^2} - 6.\left( { - 1} \right)\\
= 3 + 6 = 9\\
\to PTTT:y = 9\left( {x + 1} \right) - 3\\
\to y = 9x + 6\\
b.Có:3x + 7y - 1 = 0\\
\to y = \frac{{1 - 3x}}{7} = - \frac{3}{7}x + \frac{1}{7}\\
Do:tt \bot y = - \frac{3}{7}x + \frac{1}{7}\\
\to k.\left( { - \frac{3}{7}} \right) = - 1\\
\to k = - 1:\left( { - \frac{3}{7}} \right)\\
\to k = \frac{7}{3}\\
\to y'\left( {{x_0}} \right) = 3{x_0}^2 - 6{x_0} = \frac{7}{3}\\
\to \left[ \begin{array}{l}
{x_0} = \frac{7}{3}\\
{x_0} = - \frac{1}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
{y_0} = - \frac{{71}}{{27}}\\
{y_0} = \frac{{17}}{{27}}
\end{array} \right.\\
\to PTTT:\left[ \begin{array}{l}
y = \frac{7}{3}\left( {x - \frac{7}{3}} \right) - \frac{{71}}{{27}}\\
y = \frac{7}{3}\left( {x + \frac{1}{3}} \right) + \frac{{17}}{{27}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = \frac{7}{3}x - \frac{{218}}{{27}}\\
y = \frac{7}{3}x + \frac{{38}}{{27}}
\end{array} \right.
\end{array}\)
