Question

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Answer 1

Resposta:

Explicação passo a passo:

$a)$

$23y^{2}-[5y^{2}+29y^{2} -(25y^{2}-31y^{2})]=$

$23y^{2} -[5y^{2} +29y^{2} -25y^{2} +31y^{2} ]=$

$23y^{2} -5y^{2}-29y^{2} +25y^{2} -31y^{2}=$

$(23-5-29+25-31)y^{2} =$

$(23+25-5-29-31)y^{2} =$

$(48-65)y^{2} =$

 $-17y^{2}$

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$b)$

  $Obs.:\:mmc(2;4)=4$

$-mn+\frac{3}{4}mn-5mn+\frac{1}{2}mn=$

$-\frac{4}{4}mn +\frac{3}{4}mn-\frac{20}{4}mn+\frac{2}{4}mn=$

  $(\frac{-4\:+3\:-20\:+\:2}{4})mn=$

  $(\frac{+3\:+2\:-\:4-\:20}{4})mn=$

  $(\frac{+5\:-\:24}{4})mn=$

    $-\frac{19}{4}mn$

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$c)$

  $(3a^{2}b)\:.\:(-2bc^{3})\:.\:(-12a^{2}c^{2} )=+72a^{4}b^{2}c^{5}$

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$d)$

$(27a^{6}b^{5}c):(-3a^{2} b^{5})= -9a^{4}c$

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$e)$

  $(-5a^{3}b^{7}) ^{3} =-125a^{9}b^{21}$

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$f)$

  $(\frac{-6m^{3}n^{2} }{15}):(\frac{-9mn^{2} }{10})=$

  $(\frac{-6m^{3}n^{2} }{15})\times(\frac{10 }{-9mn^{2}})=$

      $(\frac{-60m^{3}n^{2} }{-135mn^{2} } )=$

$Vamos\;simplificar\:pelo\:mdc(60;135)=15$

    $Lembrando\;antes\:que\:(-):(-)=(+)$

           $\frac{4m^{2} }{9}$

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$Regras...$

$a^{m}\times\:a^{n}=a^{(m+n)}$

$a^{m}\div\:a^{n}=a^{(m-n)}$.

$(a^{m})^{n}=a^{m\:.\:n}$

$At\acute{e}\:Breve!$

$Jean318$