Đáp án:
$\begin{array}{l}
a){x^4}{y^4} + 4\\
= {x^4}{y^4} + 4{x^2}{y^2} + 4 - 4{x^2}{y^2}\\
= {\left( {{x^2}{y^2} + 2} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {{x^2}{y^2} + 2 + 2xy} \right)\left( {{x^2}{y^2} + 2 - 2xy} \right)\\
b)4{x^4}{y^4} + 1\\
= 4{x^4}{y^4} + 4{x^2}{y^2} + 1 - 4{x^2}{y^2}\\
= {\left( {2{x^2}{y^2} + 1} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {2{x^2}{y^2} + 2xy + 1} \right)\left( {2{x^2}{y^2} - 2xy + 1} \right)\\
c){x^4} + 1\\
= {x^4} + 2{x^2} + 1 - 2{x^2}\\
= {\left( {{x^2} + 1} \right)^2} - {\left( {\sqrt 2 x} \right)^2}\\
= \left( {{x^2} + 1 + \sqrt 2 x} \right)\left( {{x^2} + 1 - \sqrt 2 x} \right)\\
d){x^8} + x + 1\\
= {x^8} + {x^7} - {x^7} + {x^6} - {x^6} + {x^5} - {x^5} + {x^4} - {x^4}\\
+ {x^3} - {x^3} + {x^2} - {x^2} + x + 1\\
= \left( {{x^8} + {x^7} + {x^6}} \right) - \left( {{x^7} + {x^6} + {x^5}} \right)\\
+ \left( {{x^5} + {x^4} + {x^3}} \right) - \left( {{x^4} + {x^3} + {x^2}} \right)\\
+ {x^2} + x + 1\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} - {x^5} + {x^3} - {x^2} + 1} \right)\\
e){x^8} + {x^7} + 1\\
= {x^8} + {x^7} - {x^6} + {x^6} + 1\\
= {x^6}\left( {{x^2} + x + 1} \right) - \left( {{x^6} - 1} \right)\\
= {x^6}\left( {{x^2} + x + 1} \right) - \left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right)\\
= {x^6}\left( {{x^2} + x + 1} \right) - \left( {{x^3} + 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} - \left( {x - 1} \right)\left( {{x^3} + 1} \right)} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^6} - {x^4} + {x^3} - x - 1} \right)\\
f){x^8} + 3{x^4} + 1\\
= {x^8} + 2.{x^4}.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{5}{4}\\
= {\left( {{x^4} + \dfrac{3}{2}} \right)^2} - \dfrac{5}{4}\\
= \left( {{x^4} + \dfrac{{3 - \sqrt 5 }}{2}} \right)\left( {{x^4} + \dfrac{{3 + \sqrt 5 }}{2}} \right)\\
g){x^4} + 4{y^4}\\
= {x^4} + 4{x^2}{y^2} + 4{y^4} - 4{x^2}{y^2}\\
= {\left( {{x^2} + 2{y^2}} \right)^2} - {\left( {2xy} \right)^2}\\
= \left( {{x^2} + 2{y^2} + 2xy} \right)\left( {{x^2} + 2{y^2} - 2xy} \right)
\end{array}$
