Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
{\rm{DK:}}\,\,\,x \ne 2\\
a,\\
A = \dfrac{{{x^4} - 16}}{{{x^4} - 4{x^3} + 8{x^2} - 16x + 16}}\\
 = \dfrac{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}{{\left( {{x^4} - 4{x^3} + 4{x^2}} \right) + \left( {4{x^2} - 16x + 16} \right)}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}{{{x^2}\left( {{x^2} - 4x + 4} \right) + 4.\left( {{x^2} - 4x + 4} \right)}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}{{\left( {{x^2} + 4} \right).\left( {{x^2} - 4x + 4} \right)}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {{x^2} + 4} \right)}}{{\left( {{x^2} + 4} \right){{\left( {x - 2} \right)}^2}}}\\
 = \dfrac{{x + 2}}{{x - 2}}\\
b,\\
x = \sqrt {36}  = 6\\
 \Rightarrow A = \dfrac{{6 + 2}}{{6 - 2}} = \dfrac{8}{4} = 2\\
c,\\
A = \dfrac{{x + 2}}{{x - 2}} = \dfrac{{\left( {x - 2} \right) + 4}}{{x - 2}} = 1 + \dfrac{4}{{x - 2}}\\
A \in Z \Leftrightarrow \dfrac{4}{{x - 2}} \in Z\\
 \Leftrightarrow \left( {x - 2} \right) \in \left\{ { \pm 1; \pm 2; \pm 4} \right\}\\
 \Rightarrow x \in \left\{ { - 2;0;1;3;4;6} \right\}\\
2,\\
a,\\
B = \dfrac{{x{y^2} + {y^2}\left( {{y^2} - x} \right) + 1}}{{{x^2}{y^4} + 2{y^4} + {x^2} + 2}}\\
 = \dfrac{{x{y^2} + {y^4} - x{y^2} + 1}}{{\left( {{x^2}{y^4} + {x^2}} \right) + \left( {2{y^4} + 2} \right)}}\\
 = \dfrac{{{y^4} + 1}}{{{x^2}\left( {{y^4} + 1} \right) + 2.\left( {{y^4} + 1} \right)}}\\
 = \dfrac{{{y^4} + 1}}{{\left( {{y^4} + 1} \right)\left( {{x^2} + 2} \right)}}\\
 = \dfrac{1}{{{x^2} + 2}}
\end{array}\)