Đồ thị hàm số y = x^4 -2mx^2 +m có 2 điểm cực trị cùng điểm D (0;-4) tạo thành 1 hình thoi . Tìm m

\[\begin{array}{l} y = {x^4} - 2m{x^2} + m\,\,\,\left( C \right)\\ \Rightarrow y' = 4{x^3} - 4mx\\ \Rightarrow y' = 0\\ \Leftrightarrow 4{x^3} - 4mx = 0\,\,\,\left( * \right) \Leftrightarrow 4x\left( {{x^2} - m} \right) = 0\\ = \Leftrightarrow \left[ \begin{array}{l} x = 0\\ {x^2} = m \end{array} \right.\\ \Rightarrow hs\,\,\,co\,\,3\,\,\,diem\,\,\,cuc\,\,tri \Leftrightarrow \left( * \right)\,\,\,co\,\,3\,\,nghiem\,\,pb \Leftrightarrow m > 0.\\ \left( * \right) \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \sqrt m \\ x = - \sqrt m \end{array} \right. \Rightarrow \left\{ \begin{array}{l} A\left( {0;\,\,m} \right)\\ B\left( {\sqrt m ;\,\,m - {m^2}} \right)\\ C\left( { - \sqrt m ;\,\,\,m - {m^2}} \right) \end{array} \right..\\ \left( C \right)\,\,\,co\,\,\,3\,\,diem\,\,\,cuc\,\,tri\,\,\,A,\,\,B,\,\,C\,\,\,\,cung\,\,\,D\left( {0; - 4} \right)\,\,tao\,\,thanh\,\,1\,\,hinh\,\,thoi\\ Ta\,\,thay\,\,A,\,\,D \in Oy \Rightarrow ABDC\,\,\,la\,\,\,hinh\,\,thoi \Leftrightarrow AB = CD\\ Ta\,\,co:\,\,\,\overrightarrow {AB} = \left( {\sqrt m ;\,\, - {m^2}} \right);\,\,\,\overrightarrow {CD} = \left( {\sqrt m ;\,\,{m^2} - m - 4} \right)\\ \Rightarrow AB = CD \Leftrightarrow m + {m^4} = m + {\left( {{m^2} - m - 4} \right)^2}\\ \Leftrightarrow {m^4} = {\left( {{m^2} - m - 4} \right)^2}\\ \Leftrightarrow \left[ \begin{array}{l} {m^2} = {m^2} - m - 4\\ - {m^2} = {m^2} - m - 4 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = - 4\,\,\,\left( {ktm} \right)\\ 2{m^2} - m - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {tm} \right)\\ m = \frac{{1 + \sqrt {33} }}{4}\,\,\,\left( {ktm} \right) \end{array} \right. \Leftrightarrow m = \frac{{1 + \sqrt {33} }}{4}.\, \end{array}\]