`A=((3x^2)/(x^2-4)-3/(x+2)+3/(2-x)):(x+3)/(x+2)`
`a) ĐKXĐ: {(x^2-4 \ne 0),( x+2 \ne 0),(2-x \ne 0),(x+3 \ne 0):} <=> {(x\ne 2),(x\ne-2),(x\ne-3):}`
Với `x \ne 2; x \ne -2; x \ne -3` ta có:
`\qquadA=(3x^2-3(x-2)-3(x+2))/((x-2)(x+2)) . (x+2)/(x+3)`
`\qquad A=(3x^2-3x+6-3x-6)/((x-2)(x+2)). (x+2)/(x+3)`
`\qquadA=(3x^2-6x)/((x-2)(x+3))`
`\qquadA=(3x(x-2))/((x-2)(x+3))`
`\qquadA=(3x)/(x+3)`
Vậy `A=(3x)/(x+3)` với `x \ne -3; x\ne 2; x\ne-2`
`b) |x-2|=4`
`<=> `\(\left[ \begin{array}{l}x-2=4\\x-2=-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=6(\text{tm})\\x=-2(\text{ktm})\end{array} \right.\)
Thay `x=6(tmđk)` vào A ta có:
`\qquad A=(3.6)/(6+3)=18/9=2`
Vậy `A=2` khi `|x-2|=4`
`c) A=(3x)/(x+3)`
`\qquad A=(3x+9-9)/(x+3)`
`\qquad A=(3(x+3)-9)/(x+3)`
`\qquad A=3-9/(x+3)`
Để `A ∈ Z`
`=> 9/(x+3) ∈ Z`
`=> 9 \vdots x+3`
Do `x∈Z=>x+3∈Ư(9)={+-1;+-3;+-9}`
`=> x∈{-2;-4;0;-6;6;-12}`
`=> x∈{0;-4;+-6;-12}`(Loại giá trị `x=-2`)
Vậy `x∈{0;-4; +-6;-12}` thì `A∈Z`
