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

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

0 bình luận về “(1/x^2+4x+4 – 1/x^2-4x+4):(1/x+2 + 1/x+2)”

  1. `(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)`

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  2. `#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)`

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