Đáp án:
$\begin{array}{l}
a){x^2} - 3x + m + 4 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow 9 - 4\left( {m + 4} \right) > 0\\
\Rightarrow 9 - 4m - 16 > 0\\
\Rightarrow 4m < - 7\\
\Rightarrow m < \dfrac{{ - 7}}{4}\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 3\\
{x_1}{x_2} = m + 4
\end{array} \right.\\
A = x_1^2x_2^2 + 2\left( {{x_1} + {x_2}} \right)\\
= {\left( {{x_1}{x_2}} \right)^2} + 2.3\\
= {\left( {m + 4} \right)^2} + 6\\
= {m^2} + 8m + 22\\
b)m < \dfrac{{ - 7}}{4}\\
\dfrac{{{x_1}}}{{{x_2}}} + \dfrac{{{x_2}}}{{{x_1}}} = 1\\
\Rightarrow \dfrac{{x_1^2 + x_2^2}}{{{x_1}{x_2}}} = 1\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} = {x_1}{x_2}\\
\Rightarrow 9 - 3.\left( {m + 4} \right) = 0\\
\Rightarrow m + 4 = 3\\
\Rightarrow m = - 1\left( {ktm} \right)
\end{array}$
Vậy ko có m thỏa mãn.
