(x+2)/(x-2) = (x+3)/(x-3) + 2/(x²-5x+6)
Df:
x-2≠0 więc x≠2
x-3≠0 więc x≠3
x²-5x+6≠0 więc Δ=(-5)²-4*1*6=25-24=1, √Δ=1, x₁≠2, x₂≠3
Df=R\{2,3}
(x+2)/(x-2) = (x+3)/(x-3) + 2/(x²-5x+6)
(x+2)/(x-2) = (x+3)/(x-3) + 2/[(x-2)(x-3)] /*(x-2)(x-3)
(x+2)(x-3)=(x+3)(x-2)+2
x²-3x+2x-6=x²-2x+3x-6+2
x²-x²-3x+2x+2x-3x-6+6-2=0
-2x-2=0
-2x=2 /:(-2)
x=(-1)
(-1)∈Df zatem jest rozwiązaniem równania