Đáp án:

$$m = 4;\,\,m = - 1$$

Giải thích các bước giải:

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

M(x, $\frac{x+2}{x-1}$ ) (dk: x khác 1)

d(M,Ox)=| $\frac{x+2}{x-1}$ |

d(M,0y)=|x|

-> 2.| $\frac{x+2}{x-1}$ | =|x|

<-> |2x+4|=|(x-1).x|

<->|2x+4|=|x ² -x|

Th1: 2x+4=x^2-x với x>-2

x=4(tmdk) hoặc x=-1 (tmdk)

Th2: -2x-4=x ²-x với x<-2

-> vô nghiệm

có 2 điểm tm