Question

Calcule as potencias e por favor colocar toda conta grato (não consegui digitar tudo)

Answer 1

Resposta:

Explicação passo a passo:

76)

a)

$36^{\frac{1}{2} } =\sqrt{36} =6$

b)

$9^{\frac{5}{2} } =\sqrt{9^{5} } =\sqrt{(3^{2})^{5} } =\sqrt{3^{10} } =3^{\frac{10}{2} } =3^{5} =243$

c)

$32^{\frac{3}{5} } =\sqrt[5]{32^{3} } =\sqrt[5]{(2^5)^{3} } }=\sqrt[5]{2^{15} } =2^{\frac{15}{5} } =2^{3} =8$

d)

$0,25^{\frac{1}{2} } =(\frac{1}{4}) ^{\frac{1}{2} } =\sqrt{\frac{1}{4} } =\frac{\sqrt{1} }{\sqrt{4} }=\frac{1}{2}$

e)

$16^{0,75} =16^{\frac{3}{4} } =\sqrt[4]{16^{3} } =\sqrt[4]{(2^{4}) ^{3} } =\sqrt[4]{2^{12} } =2^{\frac{12}{4} } =2^{3} =8$

f)

$4^{\frac{3}{2} } =\sqrt{4^{3} } =\sqrt{64} =8$

g)

$27^{\frac{2}{3} } =\sqrt[3]{27^{2} } =\sqrt[3]{(3^{3}) ^{2} }=\sqrt[3]{3^{6} } =3^{\frac{6}{3} } =3^{2} =9$

h)

$81^{-\frac{1}{2} } =\sqrt{81^{-1} } =\sqrt{\frac{1}{81} } =\frac{\sqrt{1} }{\sqrt{81} } =\frac{1}{9}$

i)

$1^{-\frac{2}{3} }=\sqrt[3]{1^{-2} } =\sqrt[3]{\frac{1}{1^{2} } } =\sqrt[3]{\frac{1}{1} } =\sqrt[3]{1} =1$

j)

$8^{0,666...} =8^{\frac{6}{9} } =(2^{3}) ^{\frac{6}{9} } =2^{\frac{18}{9} }=2^{2} =4$

k)

$(0,36)^{0,5} =(\frac{36}{100}) ^{\frac{1}{2} } =\sqrt{\frac{36}{100} } =\frac{\sqrt{36} }{\sqrt{100} } =\frac{6}{10} =\frac{3}{5}$

l)

$100^{2,5} =100^{\frac{5}{2} } =\sqrt[]{100^{5} } =\sqrt{(10^{2}) ^{5} } =\sqrt{10^{10} } =10^{\frac{10}{2} } =10^{5} =100.000$

Answer 2

Explicação passo-a-passo:

.

Boa noite,

$a)36 {}^{ \frac{1}{2} }$

$(6 {}^{2} ) {}^{ \frac{1}{2} }$

$6 {}^{2 \times \frac{1}{2} }$

$6 {}^{ \frac{2}{2} }$

$6 {}^{1}$

$6$

.

$b)9 {}^{ \frac{5}{2} }$

$(3 {}^{2} ) {}^{ \frac{5}{2} }$

$3 {}^{2 \times \frac{5}{2} }$

$3 {}^{ \frac{10}{2} }$

$3 {}^{5}$

$243$

.

$c)32 {}^{ \frac{3}{5} }$

$(2 {}^{5} ) {}^{ \frac{3}{5} }$

$2 {}^{5 \times \frac{3}{5} }$

${2}^{ \frac{15}{5} }$

$2 {}^{3}$

$8$

.

$d)0.25 {}^{ \frac{1}{2} }$

$( \frac{25}{100} ) {}^{ \frac{1}{2} }$

$( \frac{25 {}^{ \div 25} }{100 {}^{ \div 25} } ) {}^{ \frac{1}{2} }$

$\frac{1}{4} {}^{ \frac{1}{2} }$

$\sqrt{ \frac{1}{4} }$

$\frac{1}{2}$

.

$e)16 {}^{0.75}$

$(2 {}^{4} ) {}^{ \frac{75}{100} }$

$(2 {}^{4} ) {}^{ \frac{75 {}^{ \div 25} }{100 {}^{ \div 25} } }$

$(2 {}^{4} ) {}^{ \frac{3}{4} }$

$2 {}^{4 \times \frac{3}{4} }$

$2 {}^{ \frac{12}{4} }$

$2 {}^{3}$

$8$

.

$f)4 {}^{ \frac{3}{2} }$

$(2 {}^{2} ) {}^{ \frac{3}{2} }$

$2 {}^{2 \times \frac{3}{2} }$

$2 {}^{ \frac{6}{2} }$

$2 {}^{3}$

$8$

.

$g)27 {}^{ \frac{2}{3} }$

$(3 {}^{3} ) {}^{ \frac{2}{3} }$

$3 {}^{3 \times \frac{2}{3} }$

$3 {}^{ \frac{6}{3} }$

$3 {}^{2}$

$9$

.

$h)81 {}^{ - \frac{1}{2} }$

$\frac{1}{81 {}^{ \frac{1}{2} } }$

$\frac{1}{(3 {}^{4}) {}^{ \frac{1}{2} } }$

$\frac{1}{3 {}^{4 \times \frac{1}{2} } }$

$\frac{1}{3 {}^{ \frac{4}{2} } }$

$\frac{1}{3 {}^{2} }$

$\frac{1}{9}$

.

$i)1 {}^{ - \frac{2}{3} }$

$1$

Pois, um elevado a qualquer potência, é igual a 1.

.

$j)8 {}^{0,666...}$

Primeiro vou transformar 0,666... em uma fração, para ficar mais fácil.

$0,666... = x$

$0,666... = x(10)$

$6,666... = 10x$

$10x - x = 6,666 ... - 0,666...$

$9x = 6$

$x = \dfrac{6}{9}$

$x = \dfrac{6 {}^{ \div 3} }{9 {}^{ \div 3} }$

$x = \dfrac{2}{3}$

Pronto, já que a dízima periódica, já foi transformada em uma fração, vamos calcular agora!!!

$8 {}^{ \frac{2}{3} }$

$(2 {}^{3} ) {}^{ \frac{2}{3} }$

$2 {}^{3 \times \frac{2}{3} }$

$2 {}^{ \frac{6}{3} }$

$2 {}^{2}$

$4$

.

$k)(0.36) {}^{0.5}$

$( \frac{36}{100} ) {}^{ \frac{5}{10} }$

$( \frac{36 {}^{ \div 4} }{100 {}^{ \div 4} } ) {}^{ \frac{5 {}^{ \div 5} }{10 {}^{ \div 5} } }$

$( \frac{9}{25} ) {}^{ \frac{1}{2} }$

$\sqrt{ \frac{9}{25} }$

$\frac{3}{5}$

.

$l)100 {}^{2.5}$

$100 {}^{ \frac{25}{10} }$

$100 {}^{ \frac{25 {}^{ \div 5} }{10 {}^{ \div 5} } }$

$100 {}^{ \frac{5}{2} }$

$(10 {}^{2} ) {}^{ \frac{5}{2} }$

$10 {}^{2 \times \frac{5}{2} }$

$10 {}^{ \frac{10}{2} }$

$10 {}^{5}$

$100000$