Question

E(x)=(2x/x+2 + 2x/6-3x+8x/x^2-4) : (4x^2-16x)/(3x^2+3x-18)
a) Aratati ca E(x)= (x+3)/(x-4)
b) Determinati valorile intregi ale lui x pt care E(x) apartine lui Z

Answer 1

E(x)=[2x/x+2 +2x/3(2-x) +8x/(x-2)(x+2)]:[4x(x-4)/(x+3)(x-2)*3 E(x)=[6x(x-2)-2x(x+2)+24x]/3(x-2)(x+2):4x(x-4)/3(x+3)(x-2) E(x)=(6x^2-12x-2x^2-4x+24x)/3(x-2)(x+2)*3(x+3)(x-2)/4x(x-4) E(x)=(4x^2+8x)/3(x+2)*3(x+3)/4x(x-4) E(x)=4x(x+2)/(x+2)*(x+3)/(x-4)*4x E(x)=(x+3)/(x-4) b) (x-4+7)/(x-4)=(x-4)/(x-4). +7/(x-4)=1+7/(x-4) Pentru a determina valorile intregi ale lui x trebuie sa determinam divizorii lui 7 D7={-7-1,1,7} x-4=-1. X=3 x-4=-7. x=-3 x-4=1. x=5 x-4=7. x=11 Valorile lui x ={-3,3,5,11}

Answer 2

E(x)=[2x/(x+2)+2x/3(2-x)+8x/(x-2)(x+2)]:4x(x-4)/(x-2)(x+3)
      = [2x/(x+2)+2x/-3(x-2)+8x/(x-2)(x+2)]·[(x-2)(x+3)/4x(x-4)]=
      =[2x(x+2)-2x/3(x-2)+8x/(x-2)(x+2)]·[(x-2)(x+3)/4x(x-4)]=
      ={[6x(x-2)-2x(x+2)+3·8x]/3(x-2)(x+2)}·[(x-2)(x+3)/4x(x-4)]= simplif.cu x-2
      =[(6x²-12x-2x²-4x+24x)/3(x+2)]·(x+3)/4x(x-4)=
      =[(4x²+8x)/3(x+2)]·[(x+3)/4x(x-4)]=
      =[4x(x+2)/3(x+2)]·[(x+3)/4x(x-4)]= simplif.cu 4x si x+2=
      =(x+3)/3(x-4)

b) cond.x≠4
x+3=0
x=-3