Đáp án:
$\begin{array}{l}
1)Khi:m = 2\\
{x^2} - 4x + 2 - m = 0\\
\Leftrightarrow {x^2} - 4x = 0\\
\Leftrightarrow x\left( {x - 4} \right) = 0\\
\Leftrightarrow x = 0;x = 4\\
Vậy\,x = 0;x = 4\,khi:m = 2\\
2){x^2} - 4x + 2 - m = 0\\
\Delta ' > 0\\
\Leftrightarrow 4 - 2 + m > 0\\
\Leftrightarrow m > - 2\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\\
{x_1}{x_2} = 2 - m
\end{array} \right.\\
a)x_1^2 + x_2^2 - 2{x_1}{x_2} = 3\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 3\\
\Leftrightarrow 16 - 4.\left( {2 - m} \right) = 3\\
\Leftrightarrow 16 - 8 + 4m = 3\\
\Leftrightarrow 4m = - 5\\
\Leftrightarrow m = - \dfrac{5}{4}\left( {tmdk} \right)\\
b)\dfrac{2}{{{x_1}}} + \dfrac{2}{{{x_2}}} = 4\left( {dk:m \ne 2} \right)\\
\Leftrightarrow \dfrac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = 2\\
\Leftrightarrow 4 = 2.\left( {2 - m} \right)\\
\Leftrightarrow 2 - m = 2\\
\Leftrightarrow m = 0\left( {tmdk} \right)\\
c)3{x_1} + {x_2} = 5\\
\Leftrightarrow 3{x_1} + 4 - {x_1} = 5\\
\Leftrightarrow {x_1} = \dfrac{1}{2}\\
\Leftrightarrow {x_1} = 4 - {x_1} = \dfrac{7}{2}\\
\Leftrightarrow \dfrac{1}{2}.\dfrac{7}{2} = 2 - m\\
\Leftrightarrow m = \dfrac{1}{4}\left( {tmdk} \right)\\
d){\left( {{x_1} - {x_2}} \right)^2} + 2{x_1}{x_2} = ??\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} + 2{x_1}{x_2} = ?\\
\Leftrightarrow 16 - 2\left( {2 - m} \right) = ?
\end{array}$
