Đáp án:
$\begin{array}{l}
a)m = - \frac{3}{2}\\
\Rightarrow {x^2} - 2.\left( { - \frac{3}{2} + 2} \right).x - \frac{3}{2} + 1 = 0\\
\Rightarrow {x^2} - x - \frac{1}{2} = 0\\
\Rightarrow {\left( {x - \frac{1}{2}} \right)^2} = \frac{1}{4} + \frac{1}{2}\\
\Rightarrow {\left( {x - \frac{1}{2}} \right)^2} = \frac{3}{4}\\
\Rightarrow x = \frac{{1 \pm \sqrt 3 }}{2}\\
b)\Delta ' = {\left( {m + 2} \right)^2} - m - 1\\
= {m^2} + 4m + 4 - m - 1\\
= {m^2} + 3m + 3\\
= {\left( {m + \frac{3}{2}} \right)^2} + \frac{3}{4} \ge \frac{3}{4} > 0\forall m
\end{array}$
=> pt luôn có 2 nghiệm phân biệt
$\begin{array}{l}
c)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m + 2} \right) = 2m + 4\\
{x_1}{x_2} = m + 1
\end{array} \right.\\
x_1^2 + x_2^2 = 8\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = 8\\
\Rightarrow {\left( {2m + 4} \right)^2} - 2.\left( {m + 1} \right) = 8\\
\Rightarrow 2.{\left( {m + 2} \right)^2} - m - 1 = 4\\
\Rightarrow 2{m^2} + 8m + 8 - m - 1 - 4 = 0\\
\Rightarrow 2{m^2} + 7m + 3 = 0\\
\Rightarrow \left( {2m + 1} \right)\left( {m + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = - \frac{1}{2}\\
m = - 3
\end{array} \right.
\end{array}$
