Đáp án:
$\begin{array}{l}
a)m = - 4\\
\Rightarrow {x^2} - 10x + 16 = 0\\
\Rightarrow {x^2} - 2x - 8x + 16 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x - 8} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 8
\end{array} \right.\\
b)\Delta ' \ge 0\\
\Rightarrow {\left( {m - 1} \right)^2} - {m^2} \ge 0\\
\Rightarrow {m^2} - 2m + 1 - {m^2} \ge 0\\
\Rightarrow m \le \frac{1}{2}\\
c)\left\{ \begin{array}{l}
\Delta ' > 0\\
2\left( {m - 1} \right) > 0\\
{m^2} > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m < \frac{1}{2}\\
m > 1\\
m \ne 0
\end{array} \right.\\
\Rightarrow m \in \emptyset
\end{array}$
Vậy ko có m để pt có 2 nghiệm cùng dương
d)
$\begin{array}{l}
m < \frac{1}{2}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\left( {m - 1} \right) = 2 - 2m\\
{x_1}{x_2} = {m^2}
\end{array} \right.\\
x_1^2 + x_2^2 = 14\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 14\\
\Rightarrow {\left( {2 - 2m} \right)^2} - 2{m^2} = 14\\
\Rightarrow 4{m^2} - 2{m^2} - 8m + 4 - 14 = 0\\
\Rightarrow {m^2} - 4m - 5 = 0\\
\Rightarrow \left( {m - 5} \right)\left( {m + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 5\left( {ktm} \right)\\
m = - 1\left( {tm} \right)
\end{array} \right.
\end{array}$
Vậy m=-1
