Đáp án:
$$m = 4;\,\,m = - 1$$
Giải thích các bước giải:
$$\eqalign{
& y = {{x + 2} \over {x - 1}} = \,\,\left( {x \ne 1} \right)\,\,\left( C \right) \cr
& M \in \left( C \right) \Rightarrow M\left( {m;{{m + 2} \over {m - 1}}} \right) \cr
& d\left( {M;Oy} \right) = \left| {{x_M}} \right| = \left| m \right| \cr
& d\left( {M;Ox} \right) = \left| {{y_M}} \right| = \left| {{{m + 2} \over {m - 1}}} \right| \cr
& Theo\,\,bai\,\,ra\,\,ta\,\,co: \cr
& d\left( {M;Oy} \right) = 2d\left( {M;Ox} \right) \cr
& \Rightarrow \left| m \right| = 2\left| {{{m + 2} \over {m - 1}}} \right| \cr
& \Leftrightarrow \left| {{m^2} - m} \right| = 2\left| {m + 2} \right| \cr
& TH1:\,\,\left\{ \matrix{
\left( {{m^2} - m} \right)\left( {m + 2} \right) \ge 0 \hfill \cr
{m^2} - m = 2m + 4 \hfill \cr} \right. \cr
& \Leftrightarrow \,\left\{ \matrix{
\left( {{m^2} - m} \right)\left( {m + 2} \right) \ge 0 \hfill \cr
{m^2} - 3m - 4 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left( {{m^2} - m} \right)\left( {m + 2} \right) \ge 0 \hfill \cr
\left[ \matrix{
m = 4\,\,\left( {tm} \right) \hfill \cr
m = - 1\,\,\left( {tm} \right) \hfill \cr} \right. \hfill \cr} \right. \cr
& TH2:\,\,\left\{ \matrix{
\left( {{m^2} - m} \right)\left( {m + 2} \right) < 0 \hfill \cr
{m^2} - m = - 2m - 4 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left( {{m^2} - m} \right)\left( {m + 2} \right) < 0 \hfill \cr
{m^2} + m + 4 = 0\,\,\left( {Vo\,\,nghiem} \right) \hfill \cr} \right. \cr
& Vay\,\,m = 4;\,\,m = - 1 \cr} $$
