Răspuns:
a) (x+2)(2x-3)=2x²+4x-3x-6=2x²+x-6
E(x)=[(x-6)/(x²-25) - x/(5-x) - 2/(x+5)]:(2x²+x-6)/(x²-25)=
=[(x-6)/(x-5)(x+5) + x/(x-5) - 2/(x+5)] ×(x-5)(x+5)(x+2)(2x-3)=
={[(x-6)+x(x+5)-2(x-5)]/(x-5)(x+5) ×(x-5)(x+5)/(x+2)(2x-3)=
=(x-6+x²+5x-2x+10)/(x+2)(2x-3)=
=(x²+4x+4)/(x+2)(2x-3)=
=(x+2)²/(x+2)(2x-3)=
=(x+2)/(2x-3)
c)(x+2)/(2x-3) si (2x-3)/(2x-3) ⇒
(2x+4)/(2x-3) si (2x-3)/(2x-3)
scadem ⇒7/2x-3⇒2x-3∈D7
D7={+1;-1;-7;+7} ⇒2x-3=1⇒x=2
2x-3=-1⇒x=1
2x-3=7⇒x=5
2x-3=-7 ⇒x=-2
sper ca te-am ajutat
