Đáp án:
b2) Không tồn tại m TMĐK
Giải thích các bước giải:
\(\begin{array}{l}
b1)Thay:m = 1\\
Pt \to {x^2} - 3x + 2 = 0\\
\to \left( {x - 1} \right)\left( {x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 1\\
x = 2
\end{array} \right.
\end{array}\)
b2) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to 4{m^2} + 4m + 1 - 4\left( {{m^2} + m} \right) > 0\\
\to 4{m^2} + 4m + 1 - 4{m^2} - 4m > 0\\
\to 1 > 0\left( {ld} \right)\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}{x_2} = {m^2} + m
\end{array} \right.\\
Do:1 < {x_1} < {x_2} < 3\\
\to \left\{ \begin{array}{l}
{x_1} - 1 > 0\\
{x_2} - 1 > 0\\
{x_1} - 3 < 0\\
{x_2} - 3 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6 > {x_1} + {x_2} > 2\\
\left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) > 0\\
\left( {{x_1} - 3} \right)\left( {{x_2} - 3} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6 > 2m + 1 > 2\\
{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 > 0\\
{x_1}{x_2} - 3\left( {{x_1} + {x_2}} \right) + 9 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{2} > m > \dfrac{1}{2}\\
{m^2} + m - 2m - 1 + 1 > 0\\
{m^2} + m - 3\left( {2m + 1} \right) + 9 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{2} > m > \dfrac{1}{2}\\
{m^2} - m > 0\\
{m^2} - 5m + 6 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{2} > m > \dfrac{1}{2}\\
m\left( {m - 1} \right) > 0\\
\left( {m - 2} \right)\left( {m - 3} \right) > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{2} > m > \dfrac{1}{2}\\
\left[ \begin{array}{l}
m > 1\\
m < 0
\end{array} \right.\\
\left[ \begin{array}{l}
m > 3\\
m < 2
\end{array} \right.
\end{array} \right.\left( {KTM} \right)\\
\to m \in \emptyset
\end{array}\)
