Cho biểu thức: \(A = \frac{{x - 3}}{x} - \frac{x}{{x - 3}} + \frac{9}{{{x^2} - 3x}}\).

a) Rút gọn A.

b) Tìm x để A = –3.

a) ĐKXĐ: \(\left\{ \begin{array}{l}x \ne 0\\x \ne 3\end{array} \right.\)

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

\(A = \frac{{\left( {x - 3} \right)\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} - \frac{{{x^2}}}{{x\left( {x - 3} \right)}} + \frac{9}{{x\left( {x - 3} \right)}}\)

\(A = \frac{{{x^2} - 6x + 9 - {x^2} + 9}}{{x\left( {x - 3} \right)}}\)

\(A = \frac{{ - 6x + 18}}{{x\left( {x - 3} \right)}}\)

\(A = \frac{{ - 6\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}}\)

\(A = \frac{{ - 6}}{x}\)

b) Để A = – 3 thì \(\frac{{ - 6}}{x} = - 3 \Leftrightarrow x = 2\).