Tìm điểm cố định của đồ thị sau: (Pm) : y= m x^2+(1-m)x+3 (Cm) : y= (m^2+1)x^2+(m^2+m) x-2m^2+2m-1

Đáp án:

$$\eqalign{ & a)\,\,\left( {0;3} \right);\,\,\left( {1;4} \right) \cr & b)\,\,\left( { - 2;3} \right) \cr} $$

Giải thích các bước giải:

$$\eqalign{ & \left( {{P_m}} \right):\,\,y = m{x^2} + \left( {1 - m} \right)x + 3 \cr & \Leftrightarrow y = m{x^2} + x - mx + 3 \cr & \Leftrightarrow m\left( {{x^2} - x} \right) + x + 3 - y = 0\,\,\forall m \cr & \Leftrightarrow \left\{ \matrix{ {x^2} - x = 0 \hfill \cr x + 3 - y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x = 0 \hfill \cr x = 1 \hfill \cr} \right. \hfill \cr x + 3 - y = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ x = 0 \hfill \cr y = 3 \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x = 1 \hfill \cr y = 4 \hfill \cr} \right. \hfill \cr} \right. \cr & \Rightarrow Diem\,\,co\,\,dinh:\,\,\left( {0;3} \right);\,\,\left( {1;4} \right) \cr & \cr & \left( {{C_m}} \right):\,\,y = \left( {{m^2} + 1} \right){x^2} + \left( {{m^2} + m} \right)x - 2{m^2} + 2m - 1 \cr & y = {m^2}{x^2} + {x^2} + {m^2}x + mx - 2{m^2} + 2m - 1 \cr & \Leftrightarrow {m^2}\left( {{x^2} + x - 2} \right) + m\left( {x + 2} \right) + {x^2} - 1 - y = 0\,\,\forall m \cr & \Leftrightarrow \left\{ \matrix{ {x^2} + x - 2 = 0 \hfill \cr x + 2 = 0 \hfill \cr {x^2} - 1 - y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = - 2 \hfill \cr {x^2} - 1 - y = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x = - 2 \hfill \cr y = 3 \hfill \cr} \right. \cr & \Rightarrow Diem\,\,co\,\,dinh:\,\,\left( { - 2;3} \right) \cr} $$