a) Thực hiện phép tính: \(B=\frac{{{4^5}{9^4} - {{2.6}^9}}}{{{2^{10}}{{.3}^8} + {6^8}.

* Hướng dẫn giải

a) \(B = \frac{{{4^5}{9^4} - {{2.6}^9}}}{{{2^{10}}{{.3}^8} + {6^8}.20}} = \frac{{{{({2^2})}^5}.{{({3^2})}^4} - 2.{{(2.3)}^9}}}{{{2^{10}}{{.3}^8} + {{(2.3)}^8}.({2^2}.5)}}\)

\(\frac{{{2^{10}}{{.3}^{12}} - {2^{10}}{3^9}}}{{{2^{10}}{{.3}^8} + {2^{10}}{3^8}.5}} = \frac{{{2^{10}}{{.3}^9}.({3^3} - 1)}}{{{2^{10}}{{.3}^8}(1 + 5)}} = \frac{{{2^{10}}{{.3}^9}.26}}{{{2^{10}}{{.3}^8}.6}} = 13\)

b) \({3^{n + 3}} + {3^{n + 1}} + {2^{n + 3}} + {2^{n + 2}} = {3^{n + 1}}({3^2} + 1) + {2^{n + 2}}(2 + 1)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {3^{n + 1}}.2.5 + {2^{n + 2}}.3 = 2.3({3^n}.5 + {2^{n + 1}}) \vdots 6\)

Vậy \({3^{n + 3}} + {3^{n + 1}} + {2^{n + 3}} + {2^{n + 2}} \vdots 6\) với mọi n là số nguyên dương.