\[\begin{array}{l} \left\{ \begin{array}{l} - 2mx + y = 5\\ mx + 3y = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} y = 5 + 2mx\\ mx + 3y = 1\,\,\,\,\,\,\,\,\,\left( 2 \right) \end{array} \right.\\ \Rightarrow \left( 2 \right) \Leftrightarrow mx + 3\left( {5 + 2mx} \right) = 1\\ \Leftrightarrow mx + 15 + 6mx = 1\\ \Leftrightarrow 7mx = - 14\,\,\,\,\left( * \right)\\ + )\,\,\,\,Voi\,\,\,m = 0\\ \Rightarrow \left( * \right) \Leftrightarrow 0x = - 14\\ \Rightarrow \left( * \right)\,\,\,\,\,VN.\\ + )\,\,\,Voi\,\,\,m \ne 0\\ \Rightarrow \left( * \right) \Leftrightarrow x = - \frac{{14}}{{7m}} = - \frac{2}{m}.\\ \Rightarrow y = = 5 + 2mx = 5 + 2m.\left( { - \frac{2}{m}} \right) = 1.\\ \Rightarrow hpt\,\,\,co\,\,\,nghiem\,\,\,duy\,\,\,nhat\,\,\,\,\left( {x;\,\,y} \right) = \left( { - \frac{2}{m};\,\,\,1} \right).\\ Hpt\,\,\,co\,\,\,nghiem\,\,\,{x^2} + {y^2} = 1\\ \Leftrightarrow {\left( { - \frac{2}{m}} \right)^2} + {1^2} = 1\\ \Leftrightarrow \frac{4}{{{m^2}}} = 0\\ \Rightarrow k\,\,\exists \,\,m\,\,\,thoa\,\,\,man\,\,\,bai\,\,toan. \end{array}\]

D=m2+3 , Dx=2m+5, Dy=5m−6

Dễ thấy D > 0 nên hệ có nghiệm duy nhất

⎧⎩⎨x=DxD=2m+5m2+3y=DyD=5m−6m2+3. Hai nghiệm này thỏa mãn x + y

7m−1m2+30 ⇔⎡⎣m7+33√2