Đáp án:
Câu 2:
$\begin{array}{l}
a){x^2} - \left( {2m + 1} \right).x + m = 0\\
\Delta = {\left( {2m + 1} \right)^2} - 4m\\
= 4{m^2} + 4m + 1 - 4m\\
= 4{m^2} + 1 > 0
\end{array}$
Vậy pt có 2 nghiệm phân biệt với mọi m
$\begin{array}{l}
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m + 1\\
{x_1}{x_2} = m
\end{array} \right.\\
x_1^2 - \left( {2m + 1} \right){x_1} + m = 0\\
\Leftrightarrow x_1^2 = \left( {2m + 1} \right){x_1} - m\\
Khi:A = x_1^2 - {x_1} + 2m{x_2} + {x_1}{x_2}\\
= \left( {2m + 1} \right){x_1} - m - {x_1} + 2m{x_2} + {x_1}{x_2}\\
= 2m{x_1} + 2m{x_2} + {x_1}{x_2} - m\\
= 2m\left( {{x_1} + {x_2}} \right) + m - m\\
= 2m.\left( {2m + 1} \right)\\
= 4{m^2} + 2m\\
= {\left( {2m} \right)^2} + 2.2m.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{4}\\
= {\left( {2m + \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\Leftrightarrow GTNN:A = - \dfrac{1}{4}\,khi:m = - \dfrac{1}{4}
\end{array}$
Câu 3:
$\begin{array}{l}
a){x^2} + \left( {m - 3} \right)x + m - 5 = 0\\
\Delta = {\left( {m - 3} \right)^2} - 4\left( {m - 5} \right)\\
= {m^2} - 6m + 9 - 4m + 20\\
= {m^2} - 10m + 29\\
= {\left( {m - 5} \right)^2} + 4 > 0
\end{array}$
Vậy pt có 2 nghiệm phân biệt với mọi m
b)
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3 - m\\
{x_1}{x_2} = m - 5
\end{array} \right.\\
A = {\left( {{x_1} - 1} \right)^2} + {\left( {{x_2} - 1} \right)^2}\\
= x_1^2 - 2{x_1} + 1 + x_2^2 - 2{x_2} + 1\\
= x_1^2 + x_2^2 - 2\left( {{x_1} + {x_2}} \right) + 2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 2\\
= {\left( {3 - m} \right)^2} - 2\left( {m - 5} \right) - 2\left( {3 - m} \right) + 2\\
= {m^2} - 6m + 9 - 2m + 10 - 6 + 2m + 2\\
= {m^2} - 6m + 9 + 6\\
= {\left( {m - 3} \right)^2} + 6 \ge 6\\
\Leftrightarrow GTNN:A = 6\\
Khi:m = 3
\end{array}$
