giúp em vs thank you các bn

Answer:

$\color{red}{a)}$

`A = (\frac{x-3}{x} - \frac{x}{x-3} + \frac{9}{x^2 - 3x}) : \frac{2x-2}{x}`       Điều kiện: $x\ne 0; x\ne 3$

`A = [\frac{(x-3)^2}{x(x-3)}-\frac{x^2}{x(x-3)}+\frac9{x(x-3)}] . \frac{x}{2(x-1)}`

`A = \frac{(x-3)^2 - x^2 +9}{x(x-3)} . \frac{x}{2(x-1)}`

`A = \frac{x^2 - 6x + 9 - x^2 + 9}{x(x-3)} . \frac{x}{2(x-1)}`

`A = \frac{-6x +18}{x(x-3)} . \frac{x}{2(x-1)}`

`A = \frac{-6(x-3)}{x(x-3)} . \frac{x}{2(x-1)}`

`A = \frac{-6}{x} . \frac{x}{2(x-1)}`

`A = \frac{-6}{2(x-1)}`

`A = \frac{-3}{x-1}`

$\color{red}{b)}$

Để `A = 2` thì `\frac{-3}{x-1} = 2`

`⇔ 2(x-1) = -3`

`⇔ x -1  = \frac{-3}{2}`

`⇔ x = \frac{-1}{2}` (thỏa mãn)

Vậy `x = -1/2` thì `A = 2`