Question

Olá, alguém poderia me ajudar em problemas de Função Exponencial?

CALCULE:
A) ³ ELEVADO A RAIZ DE 0,125=
B) 5 ELEVADO A RAIZ DE 100.000=
C) 7 ELEVADO A RAIZ DE 1=
D) RAIZ DE 2,25=
E) 4 ELEVADO A RAIZ DE 256 . RAIZ DE 25 - 16=
F) ³ ELEVADO A RAIZ DE 1 + RAIZ DE 49=
G) (³ ELEVADO A RAIZ DE 1000 - 6 ELEVADO A RAIZ DE 64)²=
H) RAIZ DE 9,61 - RAIZ DE 6,25 DIVIDIDO POR ³ ELEVADO A RAIZ DE 0,008=

Answer 1

Propriedades usadas:

$\sqrt[m]{a^{n}}=a^{(n/m)}\\\\\\\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$
__________________________

a)

$\sqrt[3]{0,125}=\sqrt[3]{\dfrac{125}{1000}}$

Se decompormos 125 em fatores primos, chegamos em 5³
Já 1000 = 10 . 10 . 10 = 10³

Logo:

$\sqrt[3]{0,125}=\sqrt[3]{\dfrac{5^{3}}{10^{3}}}\\\\\\\sqrt[3]{0,125}=\sqrt[3]{\left(\dfrac{5}{10}\right)^{3}}\\\\\\\sqrt[3]{0,125}=\left(\dfrac{5}{10}\right)^{3/3}\\\\\\\sqrt[3]{0,125}=\dfrac{5}{10}\\\\\\\boxed{\boxed{\sqrt[3]{0,125}=0,5}}$

b)

Veja que 100000 = 10 . 10 . 10 . 10 . 10 = 10⁵

Então:

$\sqrt[5]{100.000}=\sqrt[5]{10^{5}}\\\\\sqrt[5]{100.000}=10^{(5/5)}\\\\\sqrt[5]{100.000}=10^{1}\\\\\boxed{\boxed{\sqrt[5]{100.000}=10}}$

c)

$\sqrt[7]{1}$

1 elevado a qualquer número real é igual a 1. Portanto, sabemos que

$\sqrt[n]{1}=\sqrt[n]{1^{n}}=1^{(n/n)}=1^{1}=1~~(para~n~natural~maior~que~1)$

Logo:

$\sqrt[7]{1}=1$

d)

$\sqrt{2,25}=\sqrt[2]{\dfrac{225}{100}}$

Fatorando 225:

225 | 5
045 | 5
009 | 3
003 | 3
001

225 = 5 . 5 . 3 . 3 = 5² . 3² = (5 . 3)² = 15²
100 = 10 . 10 = 10²

$\sqrt{2,25}=\sqrt[2]{\dfrac{15^{2}}{10^{2}}}\\\\\\\sqrt{2,25}=\dfrac{\sqrt{15^{2}}}{\sqrt{10^{2}}}\\\\\\\sqrt{2,25}=\dfrac{15}{10}\\\\\\\boxed{\boxed{\sqrt{2,25}=1,5}}$

e)

$\sqrt[4]{256}\cdot\sqrt{25-16}=\sqrt[4]{256}\cdot\sqrt{9}$

Fatorando 256 e 9:

256 | 2
128 | 2
064 | 2
032 | 2
016 | 2
008 | 2
004 | 2
002 | 2
001

9 | 3
3 | 3
1

Logo:

256 = 2⁸
25 = 3²

$\sqrt[4]{256}\cdot\sqrt{25-16}=\sqrt[4]{2^{8}}\cdot\sqrt[2]{3^{2}}\\\\\sqrt[4]{256}\cdot\sqrt{25-16}=2^{(8/4)}\cdot3^{(2/2)}\\\\\sqrt[4]{256}\cdot\sqrt{25-16}=2^{2}\cdot3^{1}\\\\\sqrt[4]{256}\cdot\sqrt{25-16}=4\cdot3\\\\\boxed{\boxed{\sqrt[4]{256}\cdot\sqrt{25-16}=12}}$

f)

$\sqrt[3]{1}+\sqrt{49}$

Como foi visto anteriormente, 1 elevado a qualquer número real é 1

1 = 1³

Decompondo 49 em fatores primos:

49 | 7
07 | 7
01

$\sqrt[3]{1}+\sqrt{49}=\sqrt[3]{1^{3}}+\sqrt[2]{7^{2}}\\\\\sqrt[3]{1}+\sqrt{49}=1^{1}+7^{1}\\\\\sqrt[3]{1}+\sqrt{49}=1+7\\\\\boxed{\boxed{\sqrt[3]{1}+\sqrt{49}=8}}$

g)

$(\sqrt[3]{1000}-\sqrt[6]{64})^{2}$

64 | 2
32 | 2
16 | 2
08 | 2
04 | 2
02 | 2
01

64 = 2 . 2 . 2 . 2 . 2 . 2 = 2⁶
1000 = 10 . 10 . 10 = 10³

$(\sqrt[3]{1000}-\sqrt[6]{64})^{2}=(\sqrt[3]{10^{3}}-\sqrt[6]{2^{6}})^{2}\\\\(\sqrt[3]{1000}-\sqrt[6]{64})^{2}=(10-2)^{2}=8^{2}=8\cdot8\\\\\boxed{\boxed{(\sqrt[3]{1000}-\sqrt[6]{64})^{2}=64}}$

h)

$\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{\sqrt{(\frac{961}{100})}-\sqrt{(\frac{625}{100})}}{\sqrt[3]{(\frac{8}{1000})}}$

Decompondo 961, 625 e 8 em fatores primos:

961 | 31
031 | 31
001

625 | 5
125 | 5
025 | 5
005 | 5
001

8 | 2
4 | 2
2 | 2
1

961 = 31²
625 = 5⁴ = (5²)² = 25²
8 = 2³
100 = 10 . 10 = 10²
1000 = 100 . 10 = 10³

$\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{\sqrt{(\frac{31^{2}}{10^{2}})}-\sqrt{(\frac{25^{2}}{10^{2}})}}{\sqrt[3]{(\frac{2^{3}}{10^{3}})}}\\\\\\\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{(\frac{31}{10})-(\frac{25}{10})}{(\frac{2}{10})}\\\\\\\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{(\frac{31-25}{10})}{(\frac{2}{10})}\\\\\\\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{(\frac{6}{10})}{(\frac{2}{10})}$

$\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{6}{10}\cdot\dfrac{10}{2}\\\\\\\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=\dfrac{6}{2}\\\\\\\boxed{\boxed{\dfrac{\sqrt{9,61}-\sqrt{6,25}}{\sqrt[3]{0,008}}=3}}$