Đáp án: $y = \frac{3}{2}x - \frac{1}{2}$
Giải thích các bước giải:
$\begin{array}{l}
y = \sqrt {3x - 2} = {\left( {3x - 2} \right)^{\frac{1}{2}}}\\
\Rightarrow y' = \frac{1}{2}.\left( {3x - 2} \right)'.{\left( {3x - 2} \right)^{ - \frac{1}{2}}}\\
y' = \frac{3}{2}.\frac{1}{{\sqrt {3x - 2} }}\\
\left( d \right):3x - 2y + 1 = 0\\
\Rightarrow y = \frac{3}{2}x + \frac{1}{2}\\
Do:PTTT//\left( d \right)\\
\Rightarrow {y_o}' = \frac{3}{2}\\
\Rightarrow \frac{3}{2}.\frac{1}{{\sqrt {3{x_0} - 2} }} = \frac{3}{2}\\
\Rightarrow \sqrt {3{x_0} - 2} = 1\\
\Rightarrow {x_0} = 1\\
\Rightarrow PTTT:y = {y_0}'\left( {x - {x_0}} \right) + {y_0}\\
\Rightarrow y = \frac{3}{2}\left( {x - 1} \right) + 1 = \frac{3}{2}x - \frac{1}{2}
\end{array}$
