Question

Calcule as POTÊNCIAS NEGATIVAS:

(urgentee, as potências estão na imagem)​

Answer 1

Resposta:

$a)\\ 3^{-1}=\frac{1}{3} \approx 0,33...\\\\b)\\(-2)^{-1}=\frac{1}{(-2)} =-0,5\\\\c)\\-3^{-1}=-\frac{1}{3} } \approx -0,33...\\\\d)\\- (-5)^{-1}=-\frac{1}{(-5)}=0,2\\\\e)\\ 2^{-2}=\frac{1}{2^{2}}=\frac{1}{4} =0,25\\\\f)\\(-7)^{-2}=\frac{1}{(-7)^{2}} =\frac{1}{49} \approx 0,0204...\\\\g)\\-5^{-2}=-\frac{1}{5^{2}} =-\frac{1}{25} =-0,04\\\\h)\\(\frac{1}{3})^{-2}=\frac{(1^{-2})}{(3^{-2})}=\frac{1}{\frac{1}{3^{2}}} =\frac{1}{\frac{1}{9}} =9$

$i)\\(\frac{2}{3})^{-1}=\frac{2^{-1}}{3^{-1}}=\frac{3^{1}}{2^{1}}=\frac{3}{2} =1,5\\\\\\j)\\(-\frac{3}{2})^{-3}=(-1)^{-3}.(\frac{3^{-3}}{2^{-3}})=(-1).\frac{2^{3}}{3^{3}}=-\frac{8}{27} \approx-0,2963...\\\\\\k)\\-(\frac{2}{5})^{-2}=-(\frac{2^{-2}}{5^{-2}})=-(\frac{5^{2}}{2^{2}})=-(\frac{25}{4})=-6,25\\\\\\l)\\(0,1)^{-2}=(\frac{1}{10})^{-2}=(\frac{1^{-2}}{10^{-2}})=(\frac{10^{2}}{1^{2}})=100\\\\\\m)\\(0,25)^{-3}=(\frac{25}{100})^{-3}=(\frac{100}{25})^{3}=(4)^{3}=64$

$n)\\(-0,5)^{-3}=(-1)^{-3}.(\frac{5}{10})^{-3} =-1.(\frac{10}{5})^{3}=-(2)^{3}=-8\\\\\\o) (0,75)^{-2}=(\frac{75}{100})^{-2}=(\frac{100}{75})^{2}=(\frac{4}{3})^{2}=\frac{16}{9} \approx 1,77...$

Propriedades:

$(a^{m})^{n}=a^{m.n}\\\\\sqrt[n]{x^m} =x^{\frac{m}{n} }\\\\a^{m}a^{n}=a^{m+n}\\\\\frac{a^{m}}{a^{n}}=a^{m-n} \\\\a^{0}=1\\\\a^{1}=a$