A) (x+2)(2x-3)=2x²-3x+4x-6=2x²+x-6. B)aducem la acelsi numitor(x²-25). Prima fractie o amplifucam cu 1, a doua cu x+5, a treia cu x-5⇒E(x)=(x-6+x²+5x-2x+10)/(x²-25)*(x²-25)/(x+2)(2x-3). Simplifivam x²-25 ⇒E(x)=x²+4x+4/(x+2)(2x-3)=(x+2)²/(x+2)(2x-3). Simplificam si obtinemE(x)=(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