(x-1)/[x²(x+3)+(x+3)]+1/(x²-1)(x²+1)=(x+2)/[x²(x+3)-(x+3)]
(x-1)/(x+3)(x²+1)+1/(x²-1)(x²+1)=(x+2)/(x=3)(x²-1)
x≠-3;x≠-1;x≠1
(x-1)(x²-1)+(x+3)-(x+2)(x²+1)=0
x³-x-x²+1+x+3-x³-x-2x²-2=0
-3x²-x+2=0
3x²+x-2=0
D=1+24=25
x1=(-1-5)/6=-1не удов усл
x2=(-1+5)/6=2/3