Question

а) (25⁴)³: (4²+ 3²)¹¹ - (0²⁰¹⁰+1²⁰¹² + 2013⁰)=?

c) (2⁸+9²+5⁶+36²):(0⁶+4⁴+3⁴+25³+6⁴)=?
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Answer 1

Explicație pas cu pas:

a)$(25^{4})^{3} : (4^{2} + 3^{2})^{11} - (0^{2010} + 1^{2012} + 2013^{0} )$=

  $25^{12} : (16 + 9)^{11} - (0+1+1)$=

  $25^{12} : 25^{11} - (1+1) =$

  $25^{12-11} -2$=

  $25 - 2=$

  23

c) $(2^{8}+9^{2}+5^{6}+36^{2}):(0^{6} +4^{4}+3^{4}+25^{3}+6^{4}) =$

   $(2^{8}+81+5^{6}+36^{2}) : (4^{4}+81+25^{3}+6^{4}) =$

   $(256+81+15625 +1296) : (256+81+15625+1296)=$

   $1$