Question

3^2-3^2:(2^5--2^4-2^3-2^2-2^1+2^0
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Answer 1

 

$\displaystyle\bf\\Explicatie:\\\\\textbf{Puterile lui 2 au urmatoarea proprietate:}\\\\2^{n+1}=2^n\times2=2^n+2^n\\\\\textbf{Rezulta ca }2^{n+1}\textbf{ este de 2 ori mai mare decat }2^n\\\\\implies~2^{n+1}-2^n=2^n\\\\Exemple:\\\\2^{10}-2^9=2^9\\2^9-2^8=2^8\\2^5-2^4=2^4\\etc.Asta~ne~ajuta~sa~rezolvam~exercitiul~dat~fara~sa\\calculam~puterile.\\Rezulta~o~rezolvare~mai~eleganta.$

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$\displaystyle\bf\\Rezolvare:\\\\3^2-3^2:\Big(\underbrace{2^5-2^4}_{=2^4}-2^3-2^2-2^1+2^0\Big)=\\\\\\=3^2-3^2:\Big(2^4-2^3-2^2-2^1+2^0\Big)=\\\\=3^2-3^2:\Big(2^3-2^2-2^1+2^0\Big)=\\\\=3^2-3^2:\Big(2^2-2^1+2^0\Big)=\\\\=3^2-3^2:\Big(2^1+2^0\Big)=\\\\=3^2-3^2:\Big(2+1\Big)=\\\\=3^2-3^2:3=\\\\=3^2-3^{2-1}=\\\\=3^2-3^1=\\\\=9-3=\boxed{\bf6}$

 

 

 

Answer 2

9-9:(32-16-8-4-2+1)
9-9:3
9-3=6