Đáp án:
$\begin{array}{l}
a)m = 3\\
\Leftrightarrow {x^2} - 4x + 3 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 3} \right) = 0\\
\Leftrightarrow x = 1/x = 3\\
Vậy\,x = 1;x = 3\\
b){x^2} - 4x + m = 0\\
\Leftrightarrow \Delta ' > 0\\
\Leftrightarrow 4 - m > 0\\
\Leftrightarrow m < 4\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 4\\
{x_1}{x_2} = m
\end{array} \right.\\
x_1^3 + x_2^3 = 50\\
\Leftrightarrow \left( {{x_1} + {x_2}} \right)\left( {x_1^2 - {x_1}{x_2} + x_2^2} \right) = 50\\
\Leftrightarrow 4.\left[ {{{\left( {{x_1} + {x_2}} \right)}^2} - 3{x_1}{x_2}} \right] = 50\\
\Leftrightarrow 4.\left( {{4^2} - 3.m} \right) = 50\\
\Leftrightarrow 64 - 12m = 50\\
\Leftrightarrow 12m = 14\\
\Leftrightarrow m = \frac{7}{6}\left( {tm} \right)\\
Vậy\,m = \frac{7}{6}\\
c)m < 4\\
{x_1} - {x_2} = 4\\
\Leftrightarrow {\left( {{x_1} - {x_2}} \right)^2} = 16\\
\Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 16\\
\Leftrightarrow {4^2} - 4.m = 16\\
\Leftrightarrow 4 - m = 4\\
\Leftrightarrow m = 0\left( {tm} \right)\\
Vậy\,m = 0
\end{array}$
