E(x)={[ (x²+4x+3)(/x²+2x+1)• (x²+3x+2)(/x²+8x+15)] - (x+3)/(x-5) - 2(x-1)/(25-x²)]} : (x+3)/(x-5)
E(x)={[(x+1)(x+3)/(x+1)²·(x+1)(x+2)/(x+3)(x+5)]-(x+3)/(x-5) - 2(x-1)/-(x²-25)]}·
(x-5)/(x+3)=simplificam I fractie cu x+1
b)E(x)={[(x+3)/(x+1)·(x+1)(x+2)/(x+3)(x+5)]-(x+3)/(x-5) +2(x-1)/(x-5)(x+5)]}·
(x-5)/(x+3)=simplificam tot ce se poate
E(x)={(x+2)/(x+5)-(x+3)/(x-5)+2(x-1)/(x-5)(x+5)}·(x-5)/(x+3)=aducem acolada la acelasi numitor
E(x)={[(x+2)(x-5)-(x+3)(x+5)+2(x-1)]/(x-5)(x+5)}·(x-5)/(x+3)=simplificam cu x-5=
E(x)={[x²-5x+2x-10-(x²+5x+3x+15)+2x-2]/(x+5)}·1/(x+3)=
E(x)={(x²-3x-10-x²-8x-15+2x-2)/(x+5)}·1/(x+3)=
E(x)=(-9x-27)/(x+5)·1/(x+3)
E(x)=-9(x+3)/(x+5)(x+3)=simplificam cu x+3
E(x)=-9/(x+5)
a) cond x+5≠0⇒x≠-5
c) E(x)=(x-4)/(x+5)
cond.x≠-5
x-4=0
x=4
