Đáp án:
\(M(4; 3)\) hoặc \(M (-2; 1)\)
Giải thích các bước giải:
\(\begin{array}{l}y = \frac{{2x + 1}}{{x - 1}}\\TCD:\,\,\,x = 1\,\,\,\,\,({d_1})\\TCN:\,\,\,\,y = 2\,\,\,({d_2})\\M\left( {x;\frac{{2x + 1}}{{x - 1}}} \right)\\d(M;\,\,{d_1}) = 3d\left( {M;{d_2}} \right)\\\left| {x - 1} \right| = 3\left| {\frac{{2x + 1}}{{x - 1}} - 2} \right|\\ \Leftrightarrow \left| {x - 1} \right| = 3\left| {\frac{3}{{\left| {x - 1} \right|}}} \right|\\ \Leftrightarrow {(x - 1)^2} = 9\\ \Leftrightarrow \left[ \begin{array}{l}x = 4\,\,\,\,\,\,\,\, \Rightarrow y = 4\,\,\,\, \Rightarrow M(4;3)\\x = - 2\,\,\,\, \Rightarrow y = 1\,\,\,\, \Rightarrow M( - 2;1)\end{array} \right.\end{array}\)
