Tìm m để pt : x^4 - 2mx^2 + m - 3 = 0 a) có 1 nghiệm b) có 2 nghiệm c) có 3 nghiệm d) có 4 nghiệm giúp mình vs ạ huhu

$$\eqalign{ & {x^4} - 2m{x^2} + m - 3 = 0\,\,\left( 1 \right) \cr & Dat\,\,t = {x^2}\,\,\left( {t \ge 0} \right) \cr & PT:\,\,{t^2} - 2mt + m - 3 = 0\,\,\left( 2 \right) \cr & a)\,\,PT\,\,\left( 1 \right)\,\,co\,\,1\,\,nghiem \Leftrightarrow PT\left( * \right)\,\,co\,\,nghiem\,\,tm \cr & {t_1} \le {t_2} = 0 \cr & {t_2} = 0\,\,la\,\,nghiem\,\,cua\,\,\left( 2 \right) \cr & \Rightarrow {0^2} - 2m.0 + m - 3 = 0 \Leftrightarrow m = 3 \cr & \Rightarrow {t^2} - 6t = 0 \Leftrightarrow \left[ \matrix{ t = 0 \hfill \cr t = 6 > 0 \hfill \cr} \right.\,\,\left( {loai} \right) \cr & \Rightarrow m \in \emptyset \cr & b)\,\,Pt\left( 1 \right)\,\,co\,\,2\,\,nghiem \Leftrightarrow PT\left( 2 \right)\,\,co\,\,2\,\,nghiem\,\,trai\,\,dau \cr & \Leftrightarrow \left( {m - 3} \right).1 < 0 \Leftrightarrow m < 3 \cr & c)\,\,PT\left( 1 \right)\,\,co\,\,3\,\,nghiem \cr & \Leftrightarrow PT\left( 2 \right)\,\,co\,\,nghiem\,\,tm\,\,0 = {t_1} < {t_2} \cr & {t_1} = 0\,\,la\,\,nghiem\,cua\,\,\left( 2 \right) \cr & \Rightarrow {0^2} - 2m.0 + m - 3 = 0 \Leftrightarrow m = 3 \cr & \Rightarrow {t^2} - 6t = 0 \Leftrightarrow \left[ \matrix{ t = 0 \hfill \cr t = 6 > 0 \hfill \cr} \right.\,\,\left( {tm} \right) \cr & Vay\,\,m = 3 \cr & d)\,\,Pt\,\left( 1 \right)\,\,co\,\,4\,\,nghiem \cr & \Leftrightarrow \left( 2 \right)\,\,co\,\,2\,\,nghiem\,duong\,pb \cr & \Leftrightarrow \left\{ \matrix{ \Delta ' > 0 \hfill \cr S > 0 \hfill \cr P > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ {m^2} - m + 3 > 0\,\,\left( {luon\,\,dung} \right) \hfill \cr 2m > 0 \hfill \cr m - 3 > 0 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ m > 0 \hfill \cr m > 3 \hfill \cr} \right. \Leftrightarrow m > 3 \cr & Vay\,\,m > 3. \cr} $$