Respostas
Bruno Santos
a)-(-2)^-2=-(-1/2)^2=-1/4
b) -62,5
c)1
d)-1
e)1
f)-1
g)3/(-1/125)
h)-(1/6)/2
i)-(-6/324)^2=-(36/104976)
espero que eu tenha ajudado...poste as resposta depois pra conferir.
WALDO FRANCISCO ESTEVES JUNIOR
Bruno me ajudou demais e te agradeço etrenamente vara, valeu e conte comigo sempre
Andre Smaira
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a)
\[\eqalign{ & p = - {\left( { - 2} \right)^{ - 2}} \cr & p = - \dfrac{1}{{{{\left( { - 2} \right)}^2}}} \cr & p = - \dfrac{1}{4} }\]
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b)
\[\eqalign{ & p = - \dfrac{{{5^3}}}{2} \cr & p = - \dfrac{{125}}{2} }\]
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c)
\[\eqalign{ & p = - {\left( {\sqrt {32} } \right)^0} \cr & p = - 1 }\]
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d)
\[\eqalign{ & p = - {\left( { - \sqrt {17} } \right)^0} \cr & p = - \left( { - 1} \right) \cr & p = + 1 }\]
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e)
\[\eqalign{ & p = - {3000^0} \cr & p = - 1 }\]
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f)
\[\eqalign{ & p = - 1{\left( { - 1000000} \right)^0} \cr & p = - 1 \cdot \left( { - 1} \right) \cr & p = 1 }\]
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g)
\[\eqalign{ & p = {\left( { - \dfrac{{{5^{ - 3}}}}{3}} \right)^{ - 1}} \cr & p = {\left( {\dfrac{{ - \dfrac{1}{{{5^3}}}}}{3}} \right)^{ - 1}} \cr & p = {\left( {\dfrac{{\dfrac{1}{{125}}}}{3}} \right)^{ - 1}} \cr & p = \dfrac{3}{{\dfrac{1}{{125}}}} \cr & p = 375 }\]
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h)
\[\eqalign{ & p = - \dfrac{{{{\left( { - 3} \right)}^{ - 2}}}}{2} \cr & p = - \dfrac{{\dfrac{1}{{{{\left( { - 3} \right)}^2}}}}}{2} \cr & p = - \dfrac{{\dfrac{1}{9}}}{2} \cr & p = - \dfrac{1}{{18}} }\]
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i)
\[\eqalign{ & p = - {\left( { - \dfrac{{{{18}^2}}}{6}} \right)^{ - 2}} \cr & p = - {\left( { - 54} \right)^{ - 2}} \cr & p = - \left( {\dfrac{1}{{2916}}} \right) \cr & }\]
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