(1/x^2+4x+4 - 1/x^2-4x+4):(1/x+2 + 1/x+2)

Question

tuvi · Answer

`(1/(x^2+4x+4)-1/(x^2-4x+4)):(1/(x+2)+1/(x+2))` ĐK: `x \ne +-2`

`= (1/(x+2)^2-1/(x-2)^2): 2/(x+2)`

`= ((x-2)^2-(x+2)^2)/((x-2)^2.(x+2)^2).(x+2)/2`

`=((x-2-x-2)(x-2+x+2))/((x-2)^2.(x+2)^2).(x+2)/2`

`=(-4.2x)/((x-2)^2.(x+2)^2).(x+2)/2`

`=(-4x)/((x+2)(x^2-4x+4))`

`=(-4x)/(x^3-2x^2-4x+8)`

dananh · Answer

`#DyHungg`

`(1/(x^2+4x+4)-1/(x^2-4x+4)):(1/(x+2)+1/(x+2)) ĐKXĐ: x \ne ±2`

`=(1/((x+2)^2)-1/((x-2)^2)):2/(x+2)`

`=((x-2)^2-(x+2)^2)/((x-2)^2.(x+2)^2) . (x+2)/2`

`=((x-2-x-2)(x-2+x+2))/((x-2)^2.(x+2)^2) . (x+2)/2`

`=(-4.2x)/((x-2)^2.(x+2)^2) . (x+2)/2`

`=(-8x)/((x-2)^2.(x+2)^2) . (x+2)/2`

`=(-4x)/((x-2)^2.(x+2))`

`=(-4x)/((x^2-4x+4)(x+2))`

`=(-4x)/(x^3-2x^2-4x+8)`

(1/x^2+4x+4 – 1/x^2-4x+4):(1/x+2 + 1/x+2)