Đáp án:
Giải thích các bước giải:
`a)`
`E=(x/(x+2)-(x^3-8)/(x^3+8).(x^2-2x+4)/(x^2-4)):(1/(2+x)) (ĐK:x \ne +-2)`
`E=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4)).(x^2-2x+4)/((x-2)(x+2))).(x+2)`
`E=((x(x+2))/(x+2)^2-((x^2+2x+4))/((x+2)^2)).(x+2)`
`E=-4/(x+2)^2.(x+2)`
`E=-4/(x+2)`
`b)`
`E=-4/(x+2)`
Để `E>0`
`=>x+2<0`
`=>x<-2`
`c)`
Để `E in Z`
`=>(x+2) in Ư(4)={+-1,+-2,+-4}`
`=>x in {-1,-3,0,-4,2,-6}`
Do `x \ne 2`
`=>x in {-1,-3,0,-4,-6}`
