Đáp án+Giải thích các bước giải:
`A=(x/(x+2)+(x^3-8)/(x^3+8)*(x^2-2x+4)/(4-x^2)):4/(x+2)(x ne 2,x ne -2)`
`a)A=(x/(x+2)+((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((2-x)(x+2)))*(x+2)/4`
`A=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`A=((x^2+2x-x^2-2x-4)/(x+2)^2)*(x+2)/4`
`A=-4/(x+2)^2*(x+2)/4`
`A=-1/(x+2)`
`b)x^2-2x=0`
`<=>x(x-2)=0`
Vì `x ne 2=>x-2 ne 0`
`<=>x=0`
`=>A=-1/(0+2)=-1/2`
`b)A=3`
`<=>-1/(x+2)=3`
`<=>x+2=-1/3`
`<=>x=-1/3-2=-7/3(tm)`
`d)A<1`
`<=>-1/(x+2)-1<0`
`<=>(-x-3)/(x+2)<0`
`<=>(x+3)/(x+2)>0`
`<=>[({(x+3>0),(x+2>0):}),({(x+3<0),(x+2<0):}):}`
`<=>[({(x> -3),(x> -2):}),({(x<-3),(x<-2):}):}`
`<=>[(x> -2),(x<-3):}`
Kết hợp đk `=>[(x> -2,x ne 2),(x<-3):}`
`e) A.x=-x/(x+2) inZZ`
`=>x vdots x+2`
`=>2 vdots x+2`
`=>x+2 in Ư(2)={+-1,+-2}`
`=>x in {-1;-3;0;-4}.`
