ПОМОГИТЕ АЛГЕДРА 8 КЛАСС!!!

Question

natali_belska_3 · Accepted Answer

{ 1/(x-4)^2 + 2/(x^2-16) + 1/(x+4)^2 }  : 8x^3/(x^2-16)^2 = 1/2x
{ 1/(x-4) + 1/(x+4)}^2 * (x^2-16)^2 = 1/2x
[ {(x+4) + (x-4)}/ (x^2-16) ]^2 * (x^2-16)^2/8x^3 = 1/2x
4x^2/(x^2-16)^2 * (x^2-16)^2/8x^3 = 1/2x
4x^2/8x^3 * (x^2-16)^2/(x^2-16)^2 = 1/2x
1/2x = 1/2x

nadezhda_36028 · Answer

3
1/(x-3)-27/(x-3)(x^2+3x+9)-9/(x^2+3x+9)=
(x^2+3x+9-27-9x+27)/(x-3)(x^2+3x+9)=(x^2-6x+9)/(x-3)(x^2+3x+9)=
(x-3)^2/(x-3)(x^2+3x+9)=(x-3)/(x^2+3x+9)
(2x^2-6x+12x+18)/(x-3)=(2x^2+6x+18)/(-3)=2(x^2+3x+9)/(x-3)
(x-3)/(x^2+3x+9) * 2(x^2+3x+9)/(x-3)=2
значение выражения число постоянное