Đáp án:
$\begin{array}{l}
+ )\frac{{9{x^2} - 1}}{{12{x^2} + 4x}} = \frac{{{{\left( {3x} \right)}^2} - 1}}{{4x\left( {3x + 1} \right)}} = \frac{{\left( {3x + 1} \right)\left( {3x - 1} \right)}}{{4x\left( {3x + 1} \right)}} = \frac{{3x - 1}}{{4x}}\\
+ )\frac{{9{x^2} - 6x + 1}}{{12{x^2} - 4x}} = \frac{{{{\left( {3x - 1} \right)}^2}}}{{4x\left( {3x - 1} \right)}} = \frac{{3x - 1}}{{4x}}\\
Vậy\,\frac{{9{x^2} - 1}}{{12{x^2} + 4x}} = \frac{{9{x^2} - 6x + 1}}{{12{x^2} - 4x}}
\end{array}$
