Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{{4^{10}} + {8^4}}}{{{4^5} + {8^6}}} = \dfrac{{{{\left( {{2^2}} \right)}^{10}} + {{\left( {{2^3}} \right)}^4}}}{{{{\left( {{2^2}} \right)}^5} + {{\left( {{2^3}} \right)}^6}}} = \dfrac{{{2^{20}} + {2^{12}}}}{{{2^{10}} + {2^{18}}}} = \dfrac{{{2^{12}}.\left( {{2^8} + 1} \right)}}{{{2^{10}}.\left( {1 + {2^8}} \right)}} = \dfrac{{{2^{12}}}}{{{2^{10}}}} = {2^2} = 4\\
b,\\
B = \dfrac{{1 + {3^4} + {3^8} + {3^{12}}}}{{1 + {3^2} + {3^4} + {3^6} + {3^8} + {3^{10}} + {3^{12}} + {3^{14}}}}\\
= \dfrac{{\left( {1 + {3^4}} \right) + \left( {{3^8} + {3^{12}}} \right)}}{{\left( {1 + {3^4}} \right) + \left( {{3^2} + {3^6}} \right) + \left( {{3^8} + {3^{12}}} \right) + \left( {{3^{10}} + {3^{14}}} \right)}}\\
= \dfrac{{\left( {1 + {3^4}} \right) + {3^8}.\left( {1 + {3^4}} \right)}}{{\left( {1 + {3^4}} \right) + {3^2}.\left( {1 + {3^4}} \right) + {3^8}.\left( {1 + {3^4}} \right) + {3^{10}}.\left( {1 + {3^4}} \right)}}\\
= \dfrac{{\left( {1 + {3^4}} \right).\left( {1 + {3^8}} \right)}}{{\left( {1 + {3^4}} \right).\left( {1 + {3^2} + {3^8} + {3^{10}}} \right)}}\\
= \dfrac{{1 + {3^8}}}{{1 + {3^2} + {3^8} + {3^{10}}}}\\
= \dfrac{{1 + {3^8}}}{{\left( {1 + {3^8}} \right) + \left( {{3^2} + {3^{10}}} \right)}}\\
= \dfrac{{1 + {3^8}}}{{\left( {1 + {3^8}} \right) + {3^2}.\left( {1 + {3^8}} \right)}}\\
= \dfrac{{1 + {3^8}}}{{\left( {1 + {3^8}} \right).\left( {1 + {3^2}} \right)}}\\
= \dfrac{1}{{1 + {3^2}}} = \dfrac{1}{{10}}
\end{array}\)
