Question

Помогите пожалуйста

Answer 1

$\displaystyle\bf\\4+\frac{x^{2} +x-5}{x} +\frac{3x}{x^{2} +x-5} =0$

Решим методом замены :

$\displaystyle\bf\\\frac{x^{2} +x-5}{x} =m \ \ \ \Rightarrow \ \ \ \frac{x}{x^{2} +x-5} =\frac{1}{m} \\\\\\4+m+\frac{3}{m} =0\\\\\\\frac{m^{2} +4m+3}{m} =0\\\\\\\left \{ {{m^{2} +4m+3=0} \atop {m \neq 0}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}m_{1} =-1\\m_{2} =-3\end{array}\right } \atop {m\neq 0}} \right.$

$\displaystyle\bf\\1)\\\\m_{1} =-1\\\\\\\frac{x^{2} +x-5}{x} =-1\\\\\\\frac{x^{2} +x-5}{x} +1=0\\\\\\\frac{x^{2}+x-5+x }{x} =0\\\\\\\frac{x^{2} +2x-5}{x} =0\\\\\\x^{2} +2x-5=0 \ \ , \ \ x\neq 0\\\\\\D=2^{2} -4\cdot (-5)=4+20=24=(2\sqrt{6} )^{2} \\\\\\x_{1} =\frac{-2-2\sqrt{6} }{2} =-1-\sqrt{6} \\\\\\x_{2} =\frac{-2+2\sqrt{6} }{2} =\sqrt{6} -1$

$\displaystyle\bf\\2)\\\\m_{1} =-3\\\\\ \frac{x^{2} +x-5}{x} =-3\\\\\\\frac{x^{2} +x-5}{x} +3=0\\\\\\\frac{x^{2}+x-5+3x }{x} =0\\\\\\\frac{x^{2} +4x-5}{x} =0\\\\\\x^{2} +4x-5=0 \ \ , \ \ x\neq 0\\\\\\D=4^{2} -4\cdot (-5)=16+20=36=6^{2} \\\\\\x_{1} =\frac{-4-6}{2} =-5\\\\\\x_{2} =\frac{-4+6}{2} =1\\\\\\Otvet \ : \ -1-\sqrt{6} \ ; \ \sqrt{6} -1 \ ; \ -5 \ ; \ 1$