`a)` ĐKXĐ : ` x \ne ± 3`

` A = ( (x^2-3)/(x^2-9) +1/(x-3) ) : x/(x+3)`

` = ( (x^2-3)/((x-3)(x+3))  + (x+3)/((x+3)(x-3)) ) : x/(x+3)`

` = (x^2 - 3 + x + 3)/((x-3)(x+3)) . (x+3)/x`

`= (x^2+x)/((x-3)(x+3)) . (x+3)/x`

` = (x(x+1)(x+3))/((x-3)(x+3)x)`

` = (x+1)/(x-3)`

`b)`

` |A| = 3 \to |(x+1)/(x-3)| = 3`

Trường hợp `1`

` (x+1)/(x+3) = 3 `

`\to x + 1 = 3(x+3) `

`\to 3x + 9 = x +1 `

`\to 2x = -8 \to x = -4` (thỏa mãn)

Trường hợp `2`

` (x+1)/(x+3) = -3 `

`\to (x+1) = -3(x+3)`

` \to -3x - 9 = x+1`

`\to -4x = 10`

`\to x = 5/2` (thỏa mãn)

Vậy ` x \in { -4; 5/2 }`

`a)`

`A=((x^2-3)/(x^2-9)+1/(x-3)):x/(x+3)`

`=((x^2-3+x+3)/(x^2-9)) * (x+3)/x`

`=(x^2+x)/((x+3)(x-3)) *(x+3)/x`

`=((x^2+x)(x+3))/(x(x-3)(x+3))`

`=(x^2+x)/(x(x-3))`

`=(x(x+1))/(x(x-3))`

`=(x+1)/(x-3)`

``

`b)`

`|A|=3\to` \(\left[ \begin{array}{l}A=3\\A=-3\end{array} \right.\) 

``

Để `A=3\to(x+1)/(x-3)=3`

`\to x+1=3(x-3)`

`\to x+1=3x-9`

`\to x-3x=-9-1`

`\to -2x=-10`

`\to x=5` (thoả mãn)

``

Để `A=-3\to (x+1)/(x-3)=-3`

`\to x+1=-3x+9`

`\to x+3x=9-1`

`\to4x=8`

`\tox=2` (thoả mãn)

Vậy `x\in{2;5}` thì `|A|=3`