Đáp án:
$\begin{array}{l}
B3)\\
d)3{x^2} - 3{y^2} - 2{\left( {x - y} \right)^2}\\
= 3\left( {{x^2} - {y^2}} \right) - 2{\left( {x - y} \right)^2}\\
= 3\left( {x + y} \right)\left( {x - y} \right) - 2{\left( {x - y} \right)^2}\\
= \left( {x - y} \right)\left( {3x + 3y - 2\left( {x - y} \right)} \right)\\
= \left( {x - y} \right)\left( {3x + 3y - 2x + 2y} \right)\\
= \left( {x - y} \right)\left( {x + 5y} \right)\\
e){x^3} - 4{x^2} - 9x + 36\\
= {x^2}\left( {x - 4} \right) - 9\left( {x - 4} \right)\\
= \left( {x - 4} \right)\left( {{x^2} - 9} \right)\\
= \left( {x - 4} \right)\left( {x - 3} \right)\left( {x + 3} \right)\\
f){x^2} - {y^2} - 2x - 2y\\
= \left( {x - y} \right)\left( {x + y} \right) - 2\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - y - 2} \right)\\
B4)\\
d){\left( {x - 5} \right)^2} + \left( {x + 5} \right)\left( {x - 5} \right)\\
- \left( {5 - x} \right)\left( {2x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {x - 5 + x + 5 + 2x + 1} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {4x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - \dfrac{1}{4}
\end{array} \right.\\
Vậy\,x = 5;x = - \dfrac{1}{4}\\
e)\\
\left( {3x - 2} \right)\left( {4x - 3} \right) - \left( {2 - 3x} \right)\left( {x - 1} \right)\\
- 2\left( {3x - 2} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow 12{x^2} - 9x - 8x + 6 - \left( {2x - 2 - 3{x^2} + 3x} \right)\\
- 2\left( {3{x^2} + 3x - 2x - 2} \right) = 0\\
\Leftrightarrow 12{x^2} - 17x + 6 - \left( { - 3{x^2} + 5x - 2} \right)\\
- 6{x^2} - 2x + 4 = 0\\
\Leftrightarrow 6{x^2} - 19x + 10 + 3{x^2} - 5x + 2 = 0\\
\Leftrightarrow 9{x^2} - 24x + 12 = 0\\
\Leftrightarrow 3{x^2} - 8x + 4 = 0\\
\Leftrightarrow 3{x^2} - 2x - 6x + 4 = 0\\
\Leftrightarrow \left( {3x - 2} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 2
\end{array} \right.\\
Vậy\,x = \dfrac{2}{3};x = 2
\end{array}$
