Đáp án:`a)A=(-(x+3))/x`
`b)x>0`
`c)x in {+-1,+-3}`
Giải thích các bước giải:
`a)A=(1/3+3/(x^2-3x)):(x^2/(27-3x^2)+1/(x+3))`
`ĐK:x \ne 0,x \ne 3,x \ne -3`
`A=(1/3+3/(x(x-3)):(x^2/(3(9-x^2))+1/(x+3))`
`A=((x(x-3))/(3x(x-3))+9/(3x(x-3))):((3(x-3))/(3(x-3)(x+3))-x^2/(3(x-3)(x+3)))`
`A=((x^2-3x+9)/(3x(x-3))):((3x-9-x^2))/(3(x-3)(x+3))`
`A=(x^2-3x+9)/(3x(x-3))*(3(x-3)(x+3))/(-(x^2-3x+9))`
`A=(-(x+3))/x`
`b)A<-1`
`<=>(-(x+3))/x<-1`
`<=>(x+3)/x>1`
`<=>(x+3)/x-1>0`
`<=>(x+3-x)/x>0`
`<=>3/x>0`
Mà `3>0`
`<=>x>0`
`c)A in ZZ`
`<=>(-(x+3))/x in ZZ`
`<=>-x-3 vdots x`
`<=>3 vdots x`
`<=>x in Ư(3)={+-1,+-3}`