Với giá trị nào của  thì hai đường thẳng \({\Delta...

C

- Hướng dẫn giải

\(\left\{ \begin{array}{l} {\Delta _1}:\left\{ \begin{array}{l} x = m + 2t\\ y = 1 + \left( {{m^2} + 1} \right)t \end{array} \right. \to A\left( {m;1} \right) \in {d_1},\,\,{{\vec u}_1} = \left( {2;{m^2} + 1} \right)\\ {\Delta _2}:\left\{ \begin{array}{l} x = 1 + mt\\ y = m + t \end{array} \right. \to {{\vec u}_2} = \left( {m;1} \right) \end{array} \right.\)

\({d_1} \equiv {d_2} \Leftrightarrow \left\{ \begin{array}{l} A \in {d_2}\\ \frac{m}{2} = \frac{1}{{{m^2} + 1}} \end{array} \right.\)

\(\Leftrightarrow \left\{ \begin{array}{l} m = 1 + mt\\ 1 = m + t\\ {m^3} + m - 2 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} m = 1 + m\left( {1 - m} \right)\\ \left( {m - 1} \right)\left( {{m^2} + m + 2} \right) = 0 \end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l} {m^2} - 1 = 0\\ m - 1 = 0 \end{array} \right. \Leftrightarrow m = 1.\)