Đáp án:
\(m = \pm {1 \over {\sqrt 2 }}\)
Giải thích các bước giải:
$$\eqalign{
& y = {x^3} - 3m{x^2} + 4{m^3} \cr
& y' = 3{x^2} - 6mx = 0 \Leftrightarrow \left[ \matrix{
x = 0 \hfill \cr
x = 2m \hfill \cr} \right. \cr
& Ham\,so\,\,co\,\,2\,\,diem\,\,cuc\,\,tri \cr
& \Leftrightarrow 2m \ne 0 \Leftrightarrow m \ne 0 \cr
& x = 0 \Rightarrow y = 4{m^3} \Rightarrow A\left( {0;4{m^3}} \right) \cr
& x = 2m \Rightarrow y = 8{m^3} - 12{m^3} + 4{m^3} = 0 \Rightarrow B\left( {2m;0} \right) \cr
& A,\,\,B\,\,dx\,\,nhau\,\,qua\,\,y = x \cr
& \Leftrightarrow \left\{ \matrix{
AB \bot \,\,\left( {y = x} \right) \hfill \cr
TD\,\,I\left( {m;2{m^3}} \right)\,\,cua\,\,AB\,\, \in y = x \hfill \cr} \right. \cr
& Ta\,\,co:\,\,\overrightarrow {AB} = \left( {2m; - 4{m^3}} \right) \cr
& \,\,y = x \Leftrightarrow x - y = 0\,\,co\,\,VTCP\,\,\overrightarrow u = \left( {1;1} \right) \cr
& \Rightarrow \overrightarrow {AB} .\overrightarrow u = 0 \Leftrightarrow 2m - 4{m^3} = 0 \Leftrightarrow 2m\left( {1 - 2{m^2}} \right) = 0 \cr
& \Leftrightarrow \left[ \matrix{
m = 0\,\,\left( {loai} \right) \hfill \cr
m = \pm {1 \over {\sqrt 2 }}\,\,\left( {tm} \right) \hfill \cr} \right. \cr
& I\left( {m;2{m^3}} \right) \in y = x \cr
& \Leftrightarrow m = 2{m^3} \Leftrightarrow m\left( {2{m^2} - 1} \right) = 0 \Leftrightarrow \left[ \matrix{
m = 0\,\,\left( {loai} \right) \hfill \cr
m = \pm {1 \over {\sqrt 2 }}\,\,\left( {tm} \right) \hfill \cr} \right. \cr
& Vay\,\,m = \pm {1 \over {\sqrt 2 }} \cr} $$
