Question

Va roooog sa ma ajutati. Ultima paranteza e la puterea -1

Answer 1

E(x)={1+[x/(x+2)-(x²+x)/(x²+4x+4)]:[(x/(x²-4)-1/(x+2)]}·[(x²+2)/(x+2)]⁻¹
E(x)={1+[x/(x+2)-(x²+x)/(x+2)(x+2)]:[(x/(x+2)(x-2)-1/(x+2)]}·[(x²+2)/(x+2)]⁻¹
E(x)={1+[x(x+2)-(x²+x)]/(x+2)(x+2)]:[(x-(x-2)/(x+2)(x-2)]}·[(x+2)/(x²+2)]
E(x)={1+[(x²+2x-x²-x)]/(x+2)(x+2)]:[(x-x+2)/(x+2)(x-2)]}·[(x+2)/(x²+2)]
E(x)={1+[x/(x+2)(x+2)]:[2/(x+2)(x-2)]}·[(x+2)/(x²+2)]
E(x)={1+[x/(x+2)(x+2)]·[(x+2)(x-2)/2]}·[(x+2)/(x²+2)] simplificam cu x+2=
E(x)={1+[x/(x+2)]·[(x-2)/2]}·[(x+2)/(x²+2)] 
E(x)={1+[x(x-2)/2(x+2)]}·[(x+2)/(x²+2)] 
E(x)={(2x+4+x²-2x)/2(x+2)]}·[(x+2)/(x²+2)]  simplif.cu x+2
E(x)={(x²+4)/2}·[1/(x²+2)] 
E(x)=(x²+4)/2(x²+2)