Đáp án:

\(m = \root 3 \of {24} \)

Giải thích các bước giải:

$$\eqalign{ & y = {2 \over 3}{x^3} - {m \over 2}{x^2} - {m^2}x + 2 \cr & y' = 2{x^2} - mx - {m^2} = 0 \cr & \Delta = {m^2} + 8{m^2} = 9{m^2} \cr & De\,\,ham\,\,so\,\,co\,\,2\,\,cuc\,\,tri \Rightarrow \Delta > 0 \Leftrightarrow m \ne 0 \cr & \Rightarrow Ham\,so\,\,co\,\,2CT \cr & \left[ \matrix{ {x_1} = {{m + 3m} \over 4} = m \Rightarrow {y_1} = - {5 \over 6}{m^3} + 2 \hfill \cr {x_2} = {{m - 3m} \over 4} = - {m \over 2} \Rightarrow {y_2} = {7 \over {24}}{m^3} + 2 \hfill \cr} \right. \cr & \Rightarrow A\left( {m; - {5 \over 6}{m^3} + 2} \right);\,\,B\left( { - {m \over 2};{7 \over {24}}{m^3} + 2} \right) \cr & \Rightarrow \overrightarrow {OA} = \left( {m; - {5 \over 6}{m^3} + 2} \right) \cr & \,\,\,\,\,\overrightarrow {OB} = \left( { - {m \over 2};{7 \over {24}}{m^3} + 2} \right) \cr & De\,\,O,A,B\,\,thang\,\,hang \cr & \Rightarrow Ton\,\,tai\,\,k \ne 0:\,\,\overrightarrow {OA} = k\overrightarrow {OB} \cr & \Rightarrow \left\{ \matrix{ m = k.{{ - m} \over 2} \hfill \cr - {5 \over 6}{m^3} + 2 = k\left( {{7 \over {24}}{m^3} + 2} \right) \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ k = - 2 \hfill \cr - {5 \over 6}{m^3} + 2 = - 2\left( {{7 \over {24}}{m^3} + 2} \right)\,\,\left( * \right) \hfill \cr} \right. \cr & \left( * \right) \Leftrightarrow - {5 \over 6}{m^3} + 2 = - {7 \over {12}}{m^3} - 4 \cr & \Leftrightarrow {1 \over 4}{m^3} = 6 \Leftrightarrow {m^3} = 24 \Leftrightarrow m = \root 3 \of {24} \,\,\left( {tm} \right) \cr} $$

\[\begin{array}{l} y = \frac{2}{3}{x^3} - \frac{m}{2}{x^2} - {m^2}x + 2\\ \Rightarrow y' = 2{x^2} - mx - {m^2}\\ \Rightarrow y' = 0 \Leftrightarrow 2{x^2} - mx - {m^2} = 0\,\,\,\left( * \right)\\ \Rightarrow hs\,co\,\,2\,\,diem\,\,cuc\,\,tri \Leftrightarrow \left( * \right)\,\,co\,\,2\,\,nghiem\,\,phan\,\,biet\\ \Leftrightarrow \Delta > 0 \Leftrightarrow {m^2} + 8{m^2} > 0 \Leftrightarrow 9{m^2} > 0 \Leftrightarrow m \ne 0.\\ Ta\,\,co:\,\,\,y = \left( {\frac{1}{3}x - \frac{m}{{12}}} \right)y' - \frac{3}{4}x - \frac{{{m^3}}}{{12}} + 2\\ \Rightarrow d:\,\,\,y = - \frac{3}{4}x - \frac{{{m^3}}}{{12}} + 2\,\,\,la\,\,duong\,\,thang\,\,di\,\,qua\,\,hai\,\,diem\,\,cuc\,\,tri\,A,\,\,B\,cua\,\,hs.\\ O,\,\,A,\,\,B\,\,thang\,\,hang\\ \Rightarrow O \in d \Rightarrow - \frac{3}{4}.0 - \frac{{{m^3}}}{{12}} + 2\,\, = 0\\ \Leftrightarrow {m^3} = 24 \Leftrightarrow m = 2\sqrt[3]{3}\,\,\left( {tm} \right). \end{array}\]