Question

32/x3-2x2-x+2 + 1/(x-1)(x-2) = 1/x+1

Answer 1

$\displaystyle \frac{32}{x^3-2x^2-x+2}+\frac{1}{(x-1)(x-2)}=\frac{1}{x+1},x\neq 2,x\neq -1,x\neq 1\\ \\\frac{32}{x^3-2x^2-x+2}+\frac{1}{(x-1)(x-2)}-\frac{1}{x+1}=0\\ \\\frac{32}{x^2(x-2)-(x-2)}+\frac{1}{(x-1)(x-2)}-\frac{1}{x+1}=0\\ \\\frac{32}{(x-2)(x^2-1)}+\frac{1}{(x-1)(x-2)}-\frac{1}{x+1}=0\\ \\\frac{32}{(x-2)(x-1)(x+1)}+\frac{1}{(x-1)(x-2)}-\frac{1}{x+1}=0\\ \\\frac{32+x+1-(x-1)(x-2)}{(x+1)(x-1)(x-2)}=0\\ \\\frac{32+x+1-(x^2-2x-x+2)}{(x+1)(x-1)(x-2)}=0$

$\displaystyle \frac{32+x+1-(x^2-3x+2)}{(x+1)(x-1)(x-2)}=0\\ \\\frac{32+x+1-x^2+3x-2}{(x+1)(x-1)(x-2)}=0\\ \\\frac{31+4x-x^2}{(x+1)(x-1)(x-2)}=0\\ \\31+4x-x^2=0\\-x^2+4x+31=0\\x^2-4x-31=0\\\\x=\frac{-(-4)б\sqrt{(-4)^2-4*1*(-31)} }{2*1}=\frac{4б140}{2}=\frac{4б2\sqrt{35} }{2}\\ \\x=2б\sqrt{35}$