Question

(3x/(x^2+2x-1))-(6x/(x^2+3x-1))=5x/(x^2+4x-1)

Answer 1

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

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


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