Đáp án:
$\begin{array}{l}
a){x^2}\left( {x + 4,5} \right) = 13,5\\
\Leftrightarrow {x^3} + 4,5{x^2} - 13,5 = 0\\
\Leftrightarrow 2{x^3} + 9{x^2} - 27 = 0\\
\Leftrightarrow 2{x^3} + 6{x^2} + 3{x^2} + 9x - 9x - 27 = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2{x^2} + 3x - 9} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2{x^2} + 6x - 3x - 9} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x + 3} \right)\left( {2x - 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
2x - 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = \frac{3}{2}
\end{array} \right.\\
b)4{\left( {x + 1} \right)^2} - 9{\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow {\left( {2\left( {x + 1} \right)} \right)^2} = {\left( {3\left( {x - 1} \right)} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2\left( {x + 1} \right) = 3\left( {x - 1} \right)\\
2\left( {x + 1} \right) = - 3\left( {x - 1} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \frac{1}{5}
\end{array} \right.\\
Vậy\,x = 5\,hoặc\,x = \frac{1}{5}
\end{array}$
