Verified answer

[tex]\displaystyle\bf\\\frac{x^{2}+4x-9 }{x} -\frac{4x}{x^{2}+4x-9 } =3\\\\\\\frac{x^{2} +4x-9}{x} =m \ \ \ \Rightarrow \ \ \ \frac{4x}{x^{2} +4x-9} =\frac{4}{m} \\\\\\m-\frac{4}{m} -3=0\\\\\\\frac{m^{2}-3m-4 }{m} =0\\\\\\m^{2} -3m-4=0 \ \ , \ \ m\neq 0\\\\D=(-3)^{2} -4\cdot (-4)=9+16=25=5^{2}\\\\\\m_{1} =\frac{3-5}{2} =-1\\\\\\m_{2} =\frac{3+5}{2} =4[/tex]

[tex]\displaystyle\bf\\1)\\\\\frac{x^{2}+4x-9 }{x} =-1\\\\\\\frac{x^{2}+4x-9 }{x} +1=0\\\\\\\frac{x^{2} +4x-9+x}{x}=0\\\\\\\frac{x^{2} +5x-9}{x} =0\\\\\\x^{2} +5x-9=0 \ \ ; \ \ x\neq 0\\\\D=5^{2} -4\cdot(-9)=25+36=61\\\\\\x_{1}=\frac{-5-\sqrt{61} }{2} \\\\\\x_{2} =\frac{-5+\sqrt{61} }{2} \\\\\\2)\\\\\frac{x^{2}+4x-9 }{x} =4\\\\\\\frac{x^{2}+4x-9 }{x} -4=0\\\\\\\frac{x^{2} +4x-9-4x}{x}=0\\\\\\\frac{x^{2} -9}{x} =0\\\\x^{2} -9=0 \ \ ; \ \ x\neq 0[/tex]

[tex]\displaystyle\bf\\(x-3)(x+3)=0\\\\\\\left[\begin{array}{ccc}x-3=0\\x+3=0\end{array}\right \ \ \ \Rightarrow \ \ \ \left[\begin{array}{ccc}x_{3} =3\\x_{4}=-3 \end{array}\right\\\\\\Otvet \ : \ \frac{-5-\sqrt{61} }{2} \ \ ; \ \ \frac{\sqrt{61}-5 }{2} \ \ ; \ \ 3 \ \ ; \ \ -3[/tex]