\(\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)
\(\Leftrightarrow\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\right):\frac{2\left(x-1\right)}{x}\)
\(\frac{\left(x-3\right)\left(x-3\right)-x^2+9}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow\frac{x^2-6x+9-x^2+9}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\frac{18-6x}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow\frac{2\left(3-x\right)}{x\left(x-3\right)}.\frac{x}{2\left(x-1\right)}\)
\(\Leftrightarrow-\left(x-1\right)\)
b) Để A=2 hay \(-\left(x-1\right)=2\)
\(\Rightarrow-x+1=2\Rightarrow x=-1\)
Vậy....
