Giải và biện luận các pt sau theo tham số m a,m^2x+3mx+1=m^2-2x b, m^2(x-1)+3mx=(m^2+3)x-1

\(\begin{array}{l} a)\,\,{m^2}x + 3mx + 1 = {m^2} - 2x\\ \Leftrightarrow \left( {{m^2} + 3m + 2} \right)x = {m^2} - 1\\ \Leftrightarrow \left( {m + 1} \right)\left( {m + 2} \right)x = \left( {m + 1} \right)\left( {m - 1} \right)\,\,\,\left( * \right)\\ + )\,\,\,Voi\,\,\,m = - 1 \Rightarrow \left( * \right) \Leftrightarrow 0x = 0\\ \Rightarrow pt\,\,co\,\,\,vo\,\,so\,\,nghiem.\\ + )\,\,Voi\,\,m = - 2 \Rightarrow \left( * \right) \Leftrightarrow 0x = 3\\ \Rightarrow pt\,\,vo\,\,nghiem.\\ + )\,\,\,Voi\,\,\,\left\{ \begin{array}{l} m \ne - 1\\ m \ne - 2 \end{array} \right. \Rightarrow \left( * \right) \Leftrightarrow \left( {m + 2} \right)x = m - 1 \Leftrightarrow x = \frac{{m - 1}}{{m + 2}}\\ \Rightarrow pt\,\,co\,\,nghiem\,\,duy\,\,nhat:\,\,x = \frac{{m - 1}}{{m + 2}}.\\ b)\,\,\,{m^2}\left( {x - 1} \right) + 3mx = \left( {{m^2} + 3} \right)x - 1\\ \Leftrightarrow {m^2}x - {m^2} + 3mx - \left( {{m^2} + 3} \right)x = - 1\\ \Leftrightarrow \left( {{m^2} + 3m - {m^2} - 3} \right) = {m^2} - 1\\ \Leftrightarrow 3\left( {m - 1} \right)x = \left( {m - 1} \right)\left( {m + 1} \right)\,\,\,\,\left( * \right)\\ + )\,\,\,Voi\,\,\,m = 1 \Rightarrow \left( * \right) \Leftrightarrow 0x = 0\\ \Rightarrow pt\,\,co\,\,vo\,\,so\,\,nghiem.\\ + )\,\,\,Voi\,\,m \ne 1 \Rightarrow \left( * \right) \Leftrightarrow 3x = m + 1 \Leftrightarrow x = \frac{{m + 1}}{3}\\ \Rightarrow pt\,\,co\,\,\,nghiem\,\,\,duy\,\,nhat\,\,\,x = \frac{{m + 1}}{3}. \end{array}\)