Đáp án:
$\begin{array}{l}
a){\left( {{a^2} + {b^2} - 5} \right)^2} - 2{\left( {ab + 2} \right)^2}\\
= \left( {{a^2} + {b^2} - 5 - \sqrt 2 ab - 2\sqrt 2 } \right).\left( {{a^2} + {b^2} - 5 + \sqrt 2 ab + 2\sqrt 2 } \right)\\
b)\\
{\left( {4{a^2} - 3a - 18} \right)^2} - {\left( {4{a^2} + 3a} \right)^2}\\
= \left( {4{a^2} - 3a - 18 - 4{a^2} - 3a} \right)\\
.\left( {4{a^2} - 3a - 18 + 4{a^2} + 3a} \right)\\
= \left( { - 6a - 18} \right).\left( {8{a^2} - 18} \right)\\
= - 6.\left( {a + 3} \right).2\left( {4{a^2} - 9} \right)\\
= - 12\left( {a + 3} \right).\left( {2a - 3} \right)\left( {2a + 3} \right)\\
c) - \left( {x + 2} \right) + 3\left( {{x^2} - 4} \right)\\
= - \left( {x + 2} \right) + 3\left( {x - 2} \right)\left( {x + 2} \right)\\
= \left( {x + 2} \right)\left( { - 1 + 3\left( {x - 2} \right)} \right)\\
= \left( {x + 2} \right).\left( { - 1 + 3x - 6} \right)\\
= \left( {x + 2} \right)\left( {3x - 7} \right)\\
d)125{a^3} - 27{b^3}\\
= \left( {5a - 3b} \right)\left( {25{a^2} + 15ab + 9{b^2}} \right)
\end{array}$
