Đáp án:
\(Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 6\\
{x_1}{x_2} = 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\dfrac{1}{2}{x^2} - 3x + 2 = 0\\
Xét:\Delta = 9 - 4.\dfrac{1}{2}.2 = 5\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt 5 }}{{2.\dfrac{1}{2}}}\\
x = \dfrac{{3 - \sqrt 5 }}{{2.\dfrac{1}{2}}}
\end{array} \right. \to \left[ \begin{array}{l}
x = 3 + \sqrt 5 \\
x = 3 - \sqrt 5
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = - \dfrac{{ - 3}}{{\dfrac{1}{2}}} = 6\\
{x_1}{x_2} = \dfrac{2}{{\dfrac{1}{2}}} = 4
\end{array} \right.
\end{array}\)
