Đáp án:
$\begin{array}{l}
b){x^2} = mx + m + 1\\
\Rightarrow {x^2} - mx - m - 1 = 0\\
\Rightarrow Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = - m - 1
\end{array} \right.
\end{array}$
2 điểm nằm khác phía đối với trục tung
=> 2 hoành độ trái dấu
$\begin{array}{l}
\Rightarrow {x_1}.{x_2} < 0\\
\Rightarrow - m - 1 < 0\\
\Rightarrow m > - 1\\
Do:2\left| {{x_1}} \right| + 3 = {x_2}\\
\Rightarrow {x_1} + 2\left| {{x_1}} \right| + 3 = m\\
+ Khi:{x_1} > 0\\
\Rightarrow 3{x_1} + 3 = m\\
\Rightarrow {x_1} = \dfrac{{m - 3}}{3}\\
\Rightarrow {x_2} = \dfrac{{2m + 3}}{3}\\
\Rightarrow \dfrac{{m - 3}}{3}.\dfrac{{2m + 3}}{3} = - m - 1\\
\Rightarrow 2{m^2} - 3m - 9 = - 9m - 9\\
\Rightarrow 2{m^2} + 6m = 0\\
\Rightarrow 2m\left( {m + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 0\\
m = - 3\left( {ktm} \right)
\end{array} \right.\\
+ Khi:{x_1} < 0\\
\Rightarrow {x_1} - 2{x_1} + 3 = m\\
\Rightarrow {x_1} = 3 - m\\
\Rightarrow {x_2} = 2\left| {{x_1}} \right| + 3 = - 2{x_1} + 3 = 2m - 6\\
\Rightarrow \left( {3 - m} \right)\left( {2m - 6} \right) = - m - 1\\
\Rightarrow - 2{m^2} + 6m + 6m - 18 + m + 1 = 0\\
\Rightarrow - 2{m^2} + 13m - 17 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = \dfrac{{13 + \sqrt {33} }}{4}\left( {tm} \right)\\
m = \dfrac{{13 - \sqrt {33} }}{4}\left( {tm} \right)
\end{array} \right.\\
Vay\,m \in \left\{ {0;\dfrac{{13 + \sqrt {33} }}{4};\dfrac{{13 - \sqrt {33} }}{4}} \right\}
\end{array}$
