x²-8x+15= x²-3x-5x+15= x(x-3)-5(x-3)= (x-3)(x-5)
[(2x-8)/(x-3)(x-5) - 1/x-3 ] : 1/ (x-5)(x+5) aducem la acelasi numitor in[ ]
[(2x-8)-(x-5)] /(x-3)(x-5) :1/(x-5)(x+5)
(2x-8-x+5) /(x-3)(x-5) : 1/ (x-5)(x+5)
(x-3)/ (x-3)(x-5) : 1/ (x-5)(x+5)
1/(x-5) * (x-5)(x+5)/1= x+5
