Решите уравнение:

$x^2+5x+\frac{5}{x}+\frac{1}{x^2}=12\\\\ x^2+2+\frac{1}{x^2}+5(x+\frac{1}{x})=12+2\\\\ x^2+2*x*\frac{1}{x}+(\frac{1}{x})^2+5(x+\frac{1}{x})=12+2\\\\ (x+\frac{1}{x})^2+5*(x+\frac{1}{x})-14=0\\\\ x+\frac{1}{x}=t\\\\ t^2+5t-14=0\\\\ t^2+7t-2t-14=0\\\\ t(t+7)-2(t+7)=0\\\\ (t-2)(t+7)=0\\\\ t-2=0\ \ or\ \ t+7=0\\\\ t=2\ \ or\ \ t=-7\\\\ x+\frac{1}{x}=2\ \ or\ \ x+\frac{1}{x}=-7\\\\ x^2-2x+1=0\ \ or\ \ x^2+7x+1=0\\\\ (x-1)^2=0\ \ or\ \ x_{1,2}=\frac{-7\pm\sqrt{45}}{2}\\\\$

$x_3=1\ \ or\ \ x_{1,2}=\frac{-7\pm3\sqrt{5}}{2}$

Ответ: $1;\ \frac{-7\pm3\sqrt{5}}{2}$