Đáp án:
Xét pt hoành độ giao điểm:
$\begin{array}{l}
{x^2} = mx - m + 1\\
\Rightarrow {x^2} - mx + m - 1 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {m^2} - 4m + 4 > 0\\
\Rightarrow {\left( {m - 2} \right)^2} > 0\\
\Rightarrow m \ne 2\\
TheoViet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
a)\left| {{x_1}} \right| + \left| {{x_2}} \right| = 4\\
\Rightarrow x_1^2 + 2\left| {{x_1}{x_2}} \right| + x_2^2 = 16\\
\Rightarrow {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + 2\left| {m - 1} \right| = 16\\
\Rightarrow {m^2} - 2\left( {m - 1} \right) + 2\left| {m - 1} \right| = 16\\
+ Khi:m > 1;m \ne 2\\
\Rightarrow {m^2} - 2m + 2 + 2\left( {m - 1} \right) = 16\\
\Rightarrow {m^2} = 16\\
\Rightarrow m = 4\left( {do:m > 1;m \ne 2} \right)\\
+ Khi:m \le 1\\
\Rightarrow {m^2} - 2m + 2 - 2\left( {m - 1} \right) = 16\\
\Rightarrow {m^2} - 4m - 12 = 0\\
\Rightarrow \left( {m - 6} \right)\left( {m + 2} \right) = 0\\
\Rightarrow m = - 2\left( {m \le 1} \right)\\
Vậy\,m = - 2\,;m = 4\\
b){x_1} = 9{x_2}\\
\Rightarrow 10{x_2} = m\\
\Rightarrow {x_2} = \dfrac{m}{{10}}\\
\Rightarrow {x_1} = \dfrac{{9m}}{{10}}\\
\Rightarrow \dfrac{m}{{10}}.\dfrac{{9m}}{{10}} = m - 1\\
\Rightarrow 9{m^2} = 100m - 100\\
\Rightarrow 9{m^2} - 100m + 100 = 0\\
\Rightarrow m = 10;m = \dfrac{{10}}{9}
\end{array}$
