Question

Dau 100de puncte+coroana
1.Comparati numerele: a)x=3^542-3^541-3^540 si y=5^361
b) x=2^2008-2^2007-2^2003 si y=3^2006-3^2005-3^2003
2.Folosind proprietatile operatiilor invatate ,calculati: 
a)2007+2008*2007-2009*2006= 
b) (111+222+...+999):111+47=
 c) (2^6-2^5)*(2^5-2^4)*(2^4-2^3):(2^4)^3= 
d) (3^6-3^5)*(3^5-3^4)*(3^4-3^3):(3^4)^3 

Answer 1

$1.~a)~x=3^{542}-3^{541}-3^{540}=3^{540}(3^2-3-1)=3^{540} \cdot 5. \\ \\ Avem~de~comparat~numerele~3^{540} \cdot 5 ~si~5^{361}. \\ \\ 3^{540}=(3^3)^{180}=27^{180} \\ 5^{360}=(5^2)^{180}=25^{180} \\ \\ Avem~succesiv:~27\ \textgreater \ 25 \Leftrightarrow 27^{180}\ \textgreater \ 25^{180} \Leftrightarrow 3^{540} \cdot 5\ \textgreater \ 5^{361} \Leftrightarrow \boxed{x\ \textgreater \ y}. \\ \\ b)~x=2^{2008}-2^{2007}-2^{2003}=2^{2003}(2^5-2^4-1)=2^{2003} \cdot 15. \\ y=3^{2006}-3^{2005}-3^{2003}=3^{2003}(3^3-3^2-1)=3^{2003} \cdot 17.$

$2^{2003}\ \textless \ 3^{2003}~si~15\ \textless \ 17 \Rightarrow 2^{2003} \cdot 15 \ \textless \ 3^{2003} \cdot 17 \Leftrightarrow \boxed{x\ \textless \ y}.$

$2.~a)~2007+2008*2007-2009*2006= \\ =2007(1+2008)-2009*2006= \\ =2007*2009-2009*2006= \\ =2009(2007-2006)= \\ =2009*1= \\ =2009.$

$b)~(111+222+...+999):111+47= \\ =111(1+2+...+9): 111+47= \\ =1+2+...+9+47= \\ = \frac{9*10}{2}+47= \\ = 9*5+47= \\ =45+47= \\ =92.$

$c)~(2^6-2^5)(2^5-2^4)(2^4-2^3):(2^4)^3= \\ =2^5*(2-1)*2^4*(2-1)*2^3*(2-1):2^{12}= \\ =2^5*1*2^4*1*2^3*1:2^{12}= \\ =2^{3+4+5}:2^{12}= \\ = 2^{12}:2^{12}= \\ =1.$

$d)~(3^6-3^5)(3^5-3^4)(3^4-3^3):(3^4)^3= \\ =3^5*(3-1)*3^4*(3-1)*3^3*(3-1):3^{12}= \\ =3^{12}*2*2*2:3^{12}= \\ =2*2*2= \\ =8.$