`\qquad A=((2x)/(x-3)+(3x^2+3)/(9-x^2)+x/(x+3)):(x-1)/(x+3)`
`a)` ĐKXĐ: `{(x-3\ne0),(9-x^2\ne0),(x+3\ne0),(x-1\ne0):}<=>{(x\ne3),(x\ne-3),(x\ne1):}`
Vậy `x\ne+-3;x\ne1` thì A xác định
`b)` Với `x\ne+-3;x\ne1` thì
`A=(2x(x+3)-(3x^2+3)+x(x-3))/((x-3)(x+3)).(x+3)/(x-1)`
`A=(2x^2+6x-3x^2-3+x^2-3x)/((x-3)(x+3)).(x+3)/(x-1)`
`A=(3x-3)/((x-3)(x-1))`
`A=(3(x-1))/((x-3)(x-1))`
`A=3/(x-3)`
Vậy `A=3/(x-3)` với `x\ne+-3;x\ne1`
`c)` Thay `x=-2/3(\text{tmđk})` vào A ta có:
$A=\dfrac{3}{\dfrac{-2}{3}-3}=\dfrac{3}{\dfrac{-11}{3}}=\dfrac{-9}{11}$
Vậy `A=-9/11` khi `x=-2/3`
