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  • Brainly User
2015-09-15T01:36:10+08:00

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The answer is 1.
n = 0

 (\frac{9 ^{2(0)-1} }{(3 ^{0+1} ) ^{0} } ) ( \frac{(81 ^{0-1}) ^{0+1}  }{(27 ^{0+2} ) ^{0-1} }

 (\frac{9 ^{-1} }{3 ^{0} } )  ( \frac{(81 ^{-1}) ^{1}  }{(27 ^{0+2}) ^{-1}  } )

 (\frac{9 ^{-1} }{1} ) ( \frac{81 ^{-1} }{27 ^{-2} } )

( \frac{1}{9}) ( \frac{27 ^{2} }{81} )

   27²   
    (9) (81)

= 729
   729

= 1

FINAL ANSWER:  1

Remember that if the exponent of a number(base) is negative, it becomes a denominator, leave its factor 1.  And if the denominator has a negative exponent, it becomes a numerator, leave its factor 1.

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