Đáp án:
$\begin{array}{l}
1)b)\Delta > 0\\
\Leftrightarrow 9{\left( {m + 1} \right)^2} - 4\left( {2{m^2} - 18} \right) > 0\\
\Leftrightarrow 9{m^2} + 18m + 9 - 8{m^2} + 72 > 0\\
\Leftrightarrow {m^2} + 18m + 81 > 0\\
\Leftrightarrow {\left( {m + 9} \right)^2} > 0\\
\Leftrightarrow m\# - 9\\
Theo\,Viet:\left\{ \begin{array}{l}
a + b = 3\left( {m + 1} \right)\\
a.b = 2{m^2} - 18
\end{array} \right.\\
\left| {a - b} \right| \le 5\\
\Leftrightarrow {\left( {a - b} \right)^2} \le 25\\
\Leftrightarrow {a^2} - 2ab + {b^2} \le 25\\
\Leftrightarrow {\left( {a + b} \right)^2} - 4ab \le 25\\
\Leftrightarrow 9{\left( {m + 1} \right)^2} - 4.\left( {2{m^2} - 18} \right) \le 25\\
\Leftrightarrow {m^2} + 18m + 81 \le 25\\
\Leftrightarrow {\left( {m + 9} \right)^2} \le 25\\
\Leftrightarrow - 5 \le m + 9 \le 5\\
\Leftrightarrow - 13 \le m \le - 4\\
Vậy\, - 13 \le m \le - 4;m\# - 9\\
2)\Delta > 0\\
\Leftrightarrow 1 - 4.2.\left( { - 7} \right) > 0\\
\Leftrightarrow 1 + 56 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{1}{2}\\
{x_1}{x_2} = \dfrac{{ - 7}}{2}
\end{array} \right.\\
P = \left| {\dfrac{{{x_1}}}{{{x_2}}} - \dfrac{{{x_2}}}{{{x_1}}}} \right|\\
= \left| {\dfrac{{x_1^2 - x_2^2}}{{{x_1}{x_2}}}} \right|\\
= \dfrac{{\left| {\left( {{x_1} - {x_2}} \right)\left( {{x_1} + {x_2}} \right)} \right|}}{{\left| {\dfrac{{ - 7}}{2}} \right|}}\\
= \dfrac{2}{7}.\left| {\dfrac{1}{2}} \right|.\left| {{x_1} - {x_2}} \right|\\
= \dfrac{2}{7}.\dfrac{1}{2}.\sqrt {{{\left( {{x_1} - {x_2}} \right)}^2}} \\
= \dfrac{1}{7}.\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2} - 4{x_1}{x_2}} \\
= \dfrac{1}{7}.\sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} - 4.\dfrac{{ - 7}}{2}} \\
= \dfrac{1}{7}.\sqrt {\dfrac{1}{4} + 12} \\
= \dfrac{1}{7}.\sqrt {\dfrac{{49}}{4}} \\
= \dfrac{1}{7}.\dfrac{7}{2}\\
= \dfrac{1}{2}
\end{array}$
