$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 1:\ \\ \{_{x+\ my\ =\ 3\ \ ( 2)}^{( m-2) x\ -\ 3y\ =\ -5\ \ \ ( 1)} \ \ ( I)\\ \ \ a.\ Với\ m=1;( I) \ trở\ thành:\{_{x+\ y\ =\ 3\ \ ( 2)}^{( -x\ -\ 3y\ =\ -5\ \ \ ( 1)}\\ \ \Rightarrow ( x;y) =( 2;1)\\ b.\ ( 2) \Rightarrow x=3-my,\ thay\ vào\ \ ( 1) :\\ ( m-2)( 3-my) -3y=-5\\ \Leftrightarrow 3m-m^{2} y-6+2my-3y=-5\\ \Leftrightarrow y\left( -m^{2} +2m-3\right) =-5+6-3m\\ \Leftrightarrow y\left( m^{2} -2m+3\right) =3m-1\\ \Leftrightarrow y=\frac{3m-1}{m^{2} -2m+3} \ \left( do:\ m^{2} -2m+3 >0\right)\\ \Rightarrow Với\ \ mọi\ m\ ( I) \ đều\ có\ nghiệm\ duy\ nhất\\ Bài\ \ 2:\ \\ \ \{_{x+my\ =4\ \ ( 2)}^{mx\ -\ y\ =\ 2\ ( 1) \ } \ \ ( I)\\ Từ\ \ ( 1) \Rightarrow y=mx-2,\ thay\ \ vào\ ( 2) :\\ x+m^{2} x-2m=4\\ \Leftrightarrow x\left( m^{2} +1\right) =2m+4\\ \Leftrightarrow x=\frac{2m+4}{m^{2} +1}\\ \Rightarrow y=\frac{2m^{2} +4m}{m^{2} +1} -2=\frac{4m-2}{m^{2} +1}\\ TH1:\ m^{2} +1=1\Leftrightarrow m=0\\ \Rightarrow x=4\ ( TM) ;\ y=-2\ ( TM)\\ TH2:m^{2} +1 >1\\ Để\ x\in \mathbb{Z} \Leftrightarrow 2m+4=m^{2} +1\\ \Leftrightarrow m=\{3;-1\} \ ( 1)\\ Để\ y\in \mathbb{Z} \Leftrightarrow 4m-2=m^{2} +1\\ \Leftrightarrow m=\{3;1\} \ ( 2)\\ Từ\ ( 1)( 2) \Rightarrow m=3 \end{array}$
