[(x-2)/(x²-4)]·[(5x+10)/(x-3)+(1)]·[(x-3)/(x+2)]=1
[(x-2)/(x-2)(x+2)]·[(5x+10+x-3)/(x-3)]·[(x-3)/(x+2)]=1 se simplifica cu (x-2) prima fractie, apoi a ii si a III fractie cu x-3
[1/(x+2)]·[(6x+7)/1]·[1/(x+2)]=1
(6x+7)/(x+2)²=1
6x+7=(x+2)²
6x+7=x²+4x+4
-x²+6x-4x+7-4=0
-x²+2x+3=0 (-1)
x²-2x-3=0
Δ=4+12=16
x₁=(2-4)/2=-1
x₂=(2+4)/2=3
