[tex]\displaystyle\bf\\\frac{x^{2}+x-5 }{x} +\frac{3x}{x^{2} +x-5} +4=0\\\\\\\frac{x^{2}+x-5 }{x}=m \ \ \ \Rightarrow \ \ \ \frac{x}{x^{2} +x-5}=\frac{1}{m} \\\\\\m+\frac{3}{m} +4=0\\\\\\\frac{m^{2}+4m+3 }{m}=0 \ \ , \ \ m\neq 0\\\\\\m^{2}+4m+3=0\\\\D=4^{2} -4\cdot 3=16-12=4=2^{2} \\\\\\m_{1} =\frac{-4-2}{2}=-3\\\\\\m_{2} =\frac{-4+2}{2} =-1[/tex]

[tex]\displaystyle\bf\\1)\\\\\frac{x^{2}+x-5 }{x} =-3 \ \ , \ \ x\neq 0\\\\\\x^{2} +x-5=-3x\\\\x^{2} +4x-5=0\\\\Teorema \ Vieta \ : \ \boxed{ x_{1} \cdot x_{2} =-5}\\\\2)\\\\\frac{x^{2}+x-5 }{x} =-1\\\\x^{2} +x-5=-x\\\\x^{2} +2x-5=0\\\\Teorema \ Vieta \ : \ \boxed{ x_{3} \cdot x_{4} =-5}\\\\\\ x_{1} \cdot x_{2}\cdot x_{3} \cdot x_{4}=-5\cdot (-5)=25\\\\Otvet \ : \ 25[/tex]