a) x∧2 -2x-15=x∧2-5x+3x-15=x(x-5)+3(x-5)=(x-5)(x+3)
b)x∧2-9=(x-3)(x+3)
deci x-3≠0⇒x≠3
x+3≠0⇒x≠-3
x-5≠0⇒x≠5
atunci x∈R\{-3,3,5}
c)[x/(x-3)(x+3) +2(x-3)/(x+3)(x-3) -(x+3)/(x-3)(x+3)]:(2x-9)/(x-5)(x+3)=
[(x+2x-6-x-3)/(x-3)(x+3)] × (x-5)(x+3)/(2x-9)=
(2x-9)/(x-3)(x+3) × (x-5)(x+3)/(2x-9)=
(x-5)/(x-3)
