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