Đáp án:

\(m = \root 3 \of 3 - 2\)

Giải thích các bước giải:

\(\eqalign{ & y = {x^4} - 2\left( {m + 2} \right){x^2} + 3{\left( {m + 1} \right)^2} \cr & y' = 4{x^3} - 4\left( {m + 2} \right)x = 0 \cr & \Leftrightarrow 4x\left[ {{x^2} - m - 2} \right] = 0 \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr {x^2} = m + 2\,\,\left( * \right) \hfill \cr} \right. \cr & De\,\,ham\,\,so\,\,co\,\,3\,\,cuc\,\,tri \cr & \Rightarrow \left( * \right)\,\,co\,\,2\,\,nghiem\,\,pb\,\,khac\,\,0 \cr & \Rightarrow m + 2 > 0 \Leftrightarrow m > - 2 \cr & Khi\,\,do\,\,y' = 0 \Leftrightarrow \left[ \matrix{ x = 0 \Rightarrow y = 3{\left( {m + 1} \right)^2} \Rightarrow A\left( {0;3{{\left( {m + 1} \right)}^2}} \right) \hfill \cr x = \sqrt {m + 2} \Rightarrow y = 2{m^2} + 2m - 1 \Rightarrow B\left( {\sqrt {m + 2} ;2{m^2} + 2m - 1} \right) \hfill \cr x = - \sqrt {m + 2} \Rightarrow y = 2{m^2} + 2m - 1 \Rightarrow C\left( { - \sqrt {m + 2} ;2{m^2} + 2m - 1} \right) \hfill \cr} \right. \cr & De\,\,thay\,\,\Delta ABC\,\,can\,\,tai\,\,A \cr & \Rightarrow \Delta ABC\,\,deu\,\, \Leftrightarrow AB = BC \cr & \Leftrightarrow A{B^2} = B{C^2} \cr & \Leftrightarrow m + 2 + {\left( {m + 2} \right)^4} = 4\left( {m + 2} \right) \cr & \Leftrightarrow {\left( {m + 2} \right)^4} - 3\left( {m + 2} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ m + 2 = 0 \hfill \cr {\left( {m + 2} \right)^3} = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ m = - 2\,\,\left( {loai} \right) \hfill \cr m + 2 = \root 3 \of 3 \hfill \cr} \right. \Leftrightarrow m = \root 3 \of 3 - 2 \cr} \)