Đáp án:
Xét pt hoành độ giao điểm:
$\begin{array}{l}
{x^2} = mx - 2m + 4\\
\Rightarrow {x^2} - mx + 2m - 4 = 0\left( * \right)\\
a)m = 1\\
\Rightarrow {x^2} - x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2 \Rightarrow y = {x^2} = 4\\
x = - 1 \Rightarrow y = {x^2} = 1
\end{array} \right.\\
\Rightarrow Giao\,diem:\left( {2;4} \right);\left( { - 1;1} \right)\\
b){x^2} - mx + 2m - 4 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {m^2} - 4.\left( {2m - 4} \right) > 0\\
\Rightarrow {m^2} - 8m + 16 > 0\\
\Rightarrow {\left( {m - 4} \right)^2} > 0\\
\Rightarrow m \ne 4\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = 2m - 4
\end{array} \right.\\
\Rightarrow x_1^2 + x_2^2\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2}\\
= {m^2} - 2.\left( {2m - 4} \right)\\
= {m^2} - 4m + 8\\
= {\left( {m - 2} \right)^2} + 4 \ge 4\forall m\\
\Rightarrow Min\left( {x_1^2 + x_2^2} \right) = 4 \Leftrightarrow m = 2\left( {tmdk} \right)
\end{array}$
Vậy m=2
