\[\begin{array}{l}
Cau\,\,\,1:\\
a)\,\,2x\left( {3x - 1} \right) - \left( {x - 3} \right)\left( {6x + 2} \right)\\
= 6{x^2} - 2x - \left( {6{x^2} + 2x - 18x - 6} \right)\\
= 6{x^2} - 2x - 6{x^2} + 16x + 6\\
= 14x + 6 = 2\left( {7x + 3} \right).\\
b)\,\,{\left( {2x + 3} \right)^2} - \left( {1 + 2x} \right)\left( {2x - 1} \right) + 3\left( {2x - 3} \right)\\
= 4{x^2} + 12x + 9 - \left( {4{x^2} - 1} \right) + 6x - 9\\
= 4{x^2} + 18x - 4{x^2} + 1 = 18x + 1.\\
c)\,\,\,{\left( {x + y - 1} \right)^2} - 2\left( {x + y - 1} \right)\left( {x + y} \right) + {\left( {x + y} \right)^2}\\
= {\left( {x + y - 1 - x - y} \right)^2} = {\left( { - 1} \right)^2} = 1.\\
Cau\,\,\,2:\\
a)\,\,\,2{x^3} - 18x = 0 \Leftrightarrow 2x\left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x = 0\\
{x^2} - 9 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \pm 3
\end{array} \right..\\
b)\,\,\,\left( {3x - 2} \right)\left( {2x + 1} \right) - 6x\left( {x + 2} \right) = 11\\
\Leftrightarrow 6{x^2} + 3x - 4x - 2 - 6{x^2} - 12x - 11 = 0\\
\Leftrightarrow - 13x - 13 = 0\\
\Leftrightarrow x = - 1.\\
c)\,\,{\left( {x - 1} \right)^3} - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 3\left( {1 - {x^2}} \right)\\
\Leftrightarrow {x^3} - 3{x^2} + 3x + 1 - {x^3} - 8 = 3 - 3{x^2}\\
\Leftrightarrow 3x = 10 \Leftrightarrow x = \frac{{10}}{3}.
\end{array}\]
