Đáp án:
$\begin{array}{l}
4)a){\left( {2x + 1} \right)^2} + {\left( {2x - 1} \right)^2} - 2.\left( {2x + 1} \right)\left( {2x - 1} \right)\\
= {\left( {2x + 1} \right)^2} - 2.\left( {2x + 1} \right)\left( {2x - 1} \right) + {\left( {2x - 1} \right)^2}\\
= {\left( {2x + 1 - 2x + 1} \right)^2}\\
= {2^2}\\
= 4\\
b)\left( {2{x^3} - 3{x^2} + 6x - 9} \right):\left( {2x - 3} \right)\\
= \left[ {\left( {2x - 3} \right).{x^2} + 3.\left( {2x - 3} \right)} \right]:\left( {2x - 3} \right)\\
= \left( {{x^2} + 3} \right)\left( {2x - 3} \right):\left( {2x - 3} \right)\\
= {x^2} + 3\\
5)a)x\left( {x + 1} \right) - x\left( {x - 3} \right) = 0\\
\Rightarrow {x^2} + x - {x^2} + 3x = 0\\
\Rightarrow 4x = 0\\
\Rightarrow x = 0\\
\text{vậy}\,x = 0\\
b){x^2} - 6x + 8 = 0\\
\Rightarrow {x^2} - 2x - 4x + 8 = 0\\
\Rightarrow x\left( {x - 2} \right) - 4\left( {x - 2} \right) = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x - 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x - 4 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 4
\end{array} \right.\\
\text{vậy}\,x = 2;x = 4
\end{array}$
