Question

(3x/(x^2+2x-/(x^2+3x-1))=5x/(x^2+4x-1)

Answer 1

$\frac{3x}{x^2+2x-1}- \frac{6x}{x^2+3x-1} =\frac{5x}{x^2+4x-1}\\\\ \frac{3x}{x^2+2x-1}- \frac{6x}{x^2+3x-1} - \frac{5x}{x^2+4x-1}=0\\\\ \frac{3x(x^4+7x^3+10x^2-7x+1)-6x(x^4+6x^3+6x^2-6x+1)-5x(x^4+5x^3+4x^2-5x+1)}{(x^2+2x-1)(x^2+3x-1)(x^2+4x-1)}=0\\\\ 3x^5+21x^4+30x^3-21x^2+3x-6x^5-36x^4-36x^3+36x^2-6x-5x^5-25x^4-20x^3+25x^2-5x=0$

$-8x^5-40x^4-26x^3+40x^2-8x=0\\\\ -2x(4x^4+20x^3+13x^2-20x+4)=0\\\\ -2x^2=0\\ x=0\\\\\\ 4x^4+20x^3+13x^2-20x+4=0\\ (x+2)(4x^3+12x^2-11x+2)=0\\ (x+2)(2x-1)(2x^2+7x-2)=0\\\\ x+2=0\\ x=-2\\\\ 2x-1=0\\ 2x=1\\ x= \frac{1}{2}\\\\ 2x^2+7x-2=0\\ D=49+16=65; \ \sqrt{D}= \sqrt{65}\\\\ x_{1/2}= \frac{-7\pm \sqrt{65} }{4}$\\\\

ответ: $x_1=0; x_2=-2; x_3= \frac{1}{2}; x_4= \frac{ \sqrt{65} -7}{4}; x_5= \frac{ -\sqrt{65} -7}{4}$