[tex]\frac{(x-3m)(x+m) }{x-6}=0\\\\x-6\neq 0\\\\x\neq 6\\\\D=R\setminus \left \{ 6 \right \}\\\\rownanie\ kwadratowe\ ma\ jeden\ pierwiastek\ gdy\ \Delta =0\\\\(x-3m)(x+m)=0\\\\x^2+mx-3mx-3m^2=0\\\\x^2 -2mx-3m^2=0[/tex]
[tex]a=1,\ \ b=-2m\ \ c=-3m^2 \\\\\Delta =b^2-4ac=(-2m)^2-4*1*(-3m^2)=4m^2+12m^2=16m^2\\\\\Delta =0\\\\16m^2=0\\\\m=0[/tex]
