Đáp án: ${y^2} - \dfrac{9}{4}y - \dfrac{9}{4} = 0$
Giải thích các bước giải:
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = - 4
\end{array} \right.\\
\Rightarrow S = {y_1} + {y_2}\\
= {x_2} + \dfrac{1}{{{x_1}}} + {x_1} + \dfrac{1}{{{x_2}}}\\
= \left( {{x_1} + {x_2}} \right) + \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}}\\
= 3 + \dfrac{3}{{ - 4}} = \dfrac{9}{4}\\
P = {y_1}.{y_2} = \left( {{x_2} + \dfrac{1}{{{x_1}}}} \right).\left( {{x_1} + \dfrac{1}{{{x_2}}}} \right)\\
= {x_1}.{x_2} + 1 + 1 + \dfrac{1}{{{x_1}{x_2}}}\\
= - 4 + 2 + \dfrac{1}{{ - 4}} = - \dfrac{9}{4}\\
\Rightarrow PT:{y^2} - Sy + P = 0\\
\Rightarrow {y^2} - \dfrac{9}{4}y - \dfrac{9}{4} = 0
\end{array}$
