Đáp án:
\(a)\left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.\)
b) \(\left[ \begin{array}{l}
m = 2\\
m = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 3\\
{\mathop{\rm P}\nolimits} t \to {3^2} - 3\left( {m + 2} \right) + 2m = 0\\
\to - m + 3 = 0\\
\to m = 3\\
Thay:m = 3\\
Pt \to {x^2} - 5x + 6 = 0\\
\to \left( {x - 2} \right)\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.
\end{array}\)
b) Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to {m^2} + 4m + 4 - 4.2m \ge 0\\
\to {m^2} - 4m + 4 \ge 0\\
\to {\left( {m - 2} \right)^2} \ge 0\left( {ld} \right)\forall m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 2\\
{x_1}{x_2} = 2m
\end{array} \right.\\
{x_1}^2 + \left( {m + 2} \right){x_2} = 12\\
\to {x_1}^2 + \left( {{x_1} + {x_2}} \right){x_2} = 12\\
\to {x_1}^2 + {x_1}{x_2} + {x_2}^2 = 12\\
\to {x_1}^2 + 2{x_1}{x_2} + {x_2}^2 - {x_1}{x_2} = 12\\
\to {\left( {{x_1} + {x_2}} \right)^2} - {x_1}{x_2} = 12\\
\to {m^2} + 4m + 4 - 2m = 12\\
\to {m^2} + 2m - 8 = 0\\
\to \left( {m - 2} \right)\left( {m + 4} \right) = 0\\
\to \left[ \begin{array}{l}
m = 2\\
m = - 4
\end{array} \right.
\end{array}\)
