Question

simplifique:
a)  (a^-1+a^-2)
          a

b) (a^2)^3.b^6(c^-1)^6
      a^6.b^-6c^-6
c)$\sqrt 64a.b^6.c^1^2$

d)$\sqrt \sqrt \sqrt[3]{(64) ^2 }$

e)$5 \sqrt{3a} -7 \sqrt{12a} + \sqrt{27a}$

f)$( \sqrt{2} - \sqrt{3} ).( \sqrt{2} + \sqrt{3} )$

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9823ddc4331103ec60d160da5799cb8e.docx

Answer 1

$\frac{ a^{-1} + a^{-2} }{a} = \frac{ \frac{1}{a} + \frac{2}{a} }{a}=$

$\frac{3}{a} :a= \frac{3}{a} . \frac{1}{a} = \frac{3}{a}$

$\frac{ (a^{2})^3.b^6(c^-^1)^6}{a^6.b^-^6.c^-^6} =$

$\frac{a^6.b^6.c^-^6}{a^6.b^-^6.c^-^6} = \frac{b^6}{b^-^6}$

$b^6: b^-^6=b^6^-^(^-^6^)=b^1^2$

c) = $8a.b^6.c^1^2$

d) $\sqrt{ \sqrt{^3 \sqrt{(64)^2} } } = \sqrt{ \sqrt{^3 \sqrt{(2^6)^2} } } =$
$\sqrt{ \sqrt{^3 \sqrt{2^1^2} } } = \sqrt[12]{2^1^2} =2$

$5 \sqrt{3a} -7 \sqrt{12a} + \sqrt{27a} = 5 \sqr \sqrt{3a} -7 \sqrt{2^2.3a} + \sqrt{3^2.3a}$
$5 \sqrt{3a}-7.2 \sqrt{3a}+3 \sqrt{3a} =(5-14+3) \sqrt{3a}=-6 \sqrt{3a}$

f) $( \sqrt{2} - \sqrt{3} ).( \sqrt{2}+ \sqrt{3})=( \sqrt{2})^2 -( \sqrt{3})^2= 2-3=-1$