(1/x^2+4x+4 – 1/x^2-4x+4):(1/x+2 + 1/x+2) 01/07/2021 Bởi Natalia (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))` 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)` Bình luận
`#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)` Bình luận
`(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)`
`#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)`