Question

exercitiul 16,va roggg,URGENT​

Answer 1

Răspuns:

Vezi rezolvarile mai jos:

Explicație pas cu pas:

$a) n = 27^{9} * 32^{11} : 2 - 16^{6} * 2 * 6^{27} \\(3^{3})^{9} * (2^{5})^{11} : 2 - (2^{4})^{6} * 2 * (2*3)^{27}\\3^{27} * 2^{55} : 2 - 2^{24} * 2 * 2^{27}*3^{27} \\3^{27} * 2^{54} - 2^{52} * 3^{27}\\3^{27} * 2^{52} (2^{2} - 1)\\3^{27} * 2^{54} * 3\\3^{28} * 2^{54}\\(3^{14})^{2} *  (2^{27})^{2} este pp$

$b) n = 2^{2011} - 2^{2010} - 2^{2009} - 2x^{2008} \\2^{2008}(2^{3} - 2^{2} -2^{1} - 1)\\2^{2008} (8 - 4 - 2 - 1)\\2^{2008} * 1\\(2^{1004})^{2} este pp$

$c) n = 3^{23} * 4^{23} -2^{21} *6^{23} \\3^{23} * (2^{2})^{23} - 2^{21} * (2*3)^{23} \\3^{23} * 2^{46} - 2^{21} * 2x^{23} * 3^{23}\\     3^{23} * 2^{46} - 2^{44} * 3^{23}\\3^{23} * 2^{44}(2^{2} - 1)\\3^{23} * 2^{44} * 3\\3^{24} * 2^{44}\\(3^{12})^{2} * (2^{22})^{2} este pp$

$d) n = 3^{2011} + 2 * 3^{2010} + 3^{2009} +3^{2008} \\ 3^{2008}(3^{3} + 2 * 3^{2} + 3^{1} + 1)\\3^{2008}(27 + 2 * 9 + 3 + 1)\\3^{2008} * 49\\(3^{1004})^{2} * 7^{2}  este pp$