[tex][81^{24}-2*(3^{5})^{19}}+8^{105}:32^{63}-3^{95}]^{2017}\\Aducem\ puterile\ ori\ la\ baza\ 3\ ori\ 2.\ Ne\ vom\ folosi\ si\ de\ formula\ (a^{b})^{c}=a^{b*c}\\((3^{4})^{24}-2*3^{95}+(2^{3})^{105}:(2^{5})^{63}-3^{95})^{2017}\\=(3^{96}-2*3^{95}+2^{315}:2^{315}-3^{95})^{2017}\\Scoatem\ factor\ comun\ 3^{95}\\(3^{95}*(3-2-1)+1)^{2017}=(3^{95}*0+1)^{2017}=1^{2017}=1[/tex]
La al doilea folosesti formala lui Gauss
[tex]A=2016+2*(1+2+3+...+2015)\\\\A=2016+2*\frac{2015*2016}{2}\\A=2016+2016*2015\\A=2016*(1+2015)\\A=2016*2016=2016^{2}\ care\ este\ patrat\ perfect.\[/tex]
