Giải thích các bước giải:
Bài 1:
a,
ĐKXĐ:
\(\begin{array}{l}
9{x^2} - 6x + 1 \ne 0\\
\Leftrightarrow {\left( {3x - 1} \right)^2} \ne 0\\
\Leftrightarrow 3x - 1 \ne 0\\
\Leftrightarrow x \ne \frac{1}{3}
\end{array}\)
\(\begin{array}{l}
b,\\
x = - 4 \Rightarrow A = \frac{{3.{{\left( { - 4} \right)}^2} - \left( { - 4} \right)}}{{9.{{\left( { - 4} \right)}^2} - 6.\left( { - 4} \right) + 1}} = \frac{{52}}{{169}} = \frac{4}{{13}}
\end{array}\)
\(\begin{array}{l}
c,\\
A = \frac{{3{x^2} - x}}{{9{x^2} - 6x + 1}} = \frac{{x\left( {3x - 1} \right)}}{{{{\left( {3x - 1} \right)}^2}}} = \frac{x}{{3x - 1}}\\
2,\\
Q = {a^3} - 3{a^2} + 3a - 1 = {\left( {a - 1} \right)^3}
\end{array}\)