Đáp án:
\(\eqalign{
& a)\,\,A\left( {2;5} \right),\,\,B\left( { - 1; - 4} \right) \cr
& b)\,\,\forall m. \cr} \)
Giải thích các bước giải:
$$\eqalign{
& a)\,\,y = {x^3} - m{x^2} + mx + 2m - 3 \cr
& \Leftrightarrow \left( { - {x^2} + x + 2} \right)m + {x^3} - y - 3 = 0\,\,\forall m \cr
& \Leftrightarrow \left\{ \matrix{
- {x^2} + x + 2 = 0 \hfill \cr
{x^3} - y - 3 = 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
\left[ \matrix{
x = 2 \hfill \cr
x = - 1 \hfill \cr} \right. \hfill \cr
{x^3} - y - 3 = 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
\left\{ \matrix{
x = 2 \hfill \cr
y = 5 \hfill \cr} \right. \hfill \cr
\left\{ \matrix{
x = - 1 \hfill \cr
y = - 4 \hfill \cr} \right. \hfill \cr} \right. \Rightarrow \left[ \matrix{
A\left( {2;5} \right) \hfill \cr
B\left( { - 1; - 4} \right) \hfill \cr} \right. \cr
& \Rightarrow DTHS\,luon\,\,di\,\,qua\,\,A\left( {2;5} \right),\,\,B\left( { - 1; - 4} \right)\,\,\forall m. \cr
& b)\,\,PTAB:\,\,{{x - 2} \over { - 1 - 2}} = {{y - 5} \over { - 4 - 5}} \cr
& \Leftrightarrow {{x - 2} \over { - 3}} = {{y - 5} \over { - 9}} \Leftrightarrow 3\left( {x - 2} \right) = y - 5 \cr
& \Leftrightarrow 3x - y - 1 = 0 \Leftrightarrow y = 3x - 1 \cr
& De\,\,AB\,\,tiep\,\,xuc\,\,voi\,\,DTHS\,\,thi\,\,hpt \cr
& \left\{ \matrix{
{x^3} - m{x^2} + mx + 2m - 3 = 3x - 1\,\,\left( 1 \right) \hfill \cr
3{x^2} - 2mx + m = 3\,\,\,\left( 2 \right) \hfill \cr} \right.\,\,co\,\,nghiem \cr
& \Rightarrow \left( 2 \right)\,\,co\,\,nghiem \cr
& \Leftrightarrow \Delta ' = {m^2} - 3\left( {m - 3} \right) = {m^2} - 3m + 9 \ge 0\,\,\left( {luon\,\,dung} \right) \cr
& Vay\,\,AB\,\,tiep\,\,xuc\,\,voi\,\,DTHS\,\,\forall m. \cr} $$
