Question

x^2+x-5/x + 3x/(x^2+x-5) + 4 =0

Answer 1

$frac{x^2+x-5}x+frac{3x}{x^2+x-5}+4=0\O.D.3.:\begin{cases}xneq0\x^2+x-5neq0end{cases}Rightarrowbegin{cases}xneq0\xneqfrac{-1-sqrt{21}}2\xneqfrac{-1+sqrt{21}}2end{cases}\\frac{(x^2+x-5)^2+3x^2+4x(x^2+x-5)}{x(x^2+x-5)}=0\(x^2+x-5)^2+3x^2+4x(x^2+x-5)=0\x^4+2x^3-9x^2-10x+25+3x^2+4x^2+4x^3-20x=0\x^4+6x^3-2x^2-30x+25=0\(x^2+4x-5)(x^2+2x-5)=0\(x+1)(x-5)(x^2+2x-5)=0\x_1=-1,;x_2=5,;x_3=-1-sqrt6,;x_4=sqrt6-1$