Giải thích các bước giải:
a,
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
x \ne 3
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
A = \frac{{x - 3}}{x} - \frac{x}{{x - 3}} + \frac{9}{{{x^2} - 3x}}\\
= \frac{{\left( {x - 3} \right)\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} - \frac{{x.x}}{{x\left( {x - 3} \right)}} + \frac{9}{{x\left( {x - 3} \right)}}\\
= \frac{{{{\left( {x - 3} \right)}^2} - {x^2} + 9}}{{x\left( {x - 3} \right)}}\\
= \frac{{{x^2} - 6x + 9 - {x^2} + 9}}{{x\left( {x - 3} \right)}}\\
= \frac{{ - 6x + 18}}{{x\left( {x - 3} \right)}}\\
= \frac{{ - 6\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}}\\
= \frac{{ - 6}}{x}
\end{array}\)
b,
\(A = - 3 \Leftrightarrow \frac{{ - 6}}{x} = - 3 \Leftrightarrow x = 2\)
