Question

Simplefique usando a propriedade conveniente:

A)
$\sqrt[8]{ {3}^{2} }$
B)
$\sqrt{ {a}^{10} }$
C)

$\sqrt[4]{b}$
D)
$\sqrt[]{ {16}^{} }$

E)
$\sqrt[3]{64}$
F)
$\sqrt[3]{27}$
G)
$\sqrt[3]{8 {a}^{6} }$
H)

$\sqrt{64x {}^{2} } {y}^{8}$
I)
$\sqrt[4]{16x {}^{8} }$
J)
$\sqrt{25 \times {a}^{2} } \times b {}^{2}$
L)

$\sqrt{75}$
M)

$\sqrt{18}$
N)

$\sqrt{50}$
O)
$\sqrt{48 {}^{a} }$
P)
$\sqrt{ \frac{9}{x {}^{2} } }$
Q)
$\sqrt{ \frac{25 \times a}{ {x}^{10} } }$
N)
$\sqrt{ \frac{49 \times \: a {}^{2} }{16} }$

Answer 1

Olá, tudo bem?

Vamos lá:
$A) \sqrt[8]{3^2}= \sqrt[8\div2]{3^{2\div2}}= \sqrt[4]{3}=\ \textgreater \ \boxed{\boxed{3^{ \frac{1}{4} } }}$

$B) \sqrt{a^{10}}= \sqrt[2\div2]{a^{10\div2}}=\ \textgreater \ \boxed{\boxed{a^5 }}\\\\\\ C) \sqrt[4]{b}=\ \textgreater \ \boxed{\boxed{b^{\frac{1}{4}} }}\\\\\\ D) \sqrt{16}= \sqrt{4^2}= \sqrt[2\div2]{4^{2\div2}}=\ \textgreater \ \boxed{\boxed{4 }}\\\\\\ E) \sqrt[3]{64}= \sqrt[3]{4^3}= \sqrt[3\div3]{4^{3\div3}}=\ \textgreater \ \boxed{\boxed{4 }}\\\\\\ F) \sqrt[3]{27}= \sqrt[3]{3^3}= \sqrt[3\div3]{3^{3\div3}}=\ \textgreater \ \boxed{\boxed{3 }}$

$G) \sqrt[3]{8a^6}= \sqrt[3]{2^3a^6}= \sqrt[3\div3]{2^{3\div3}\cdot a^{6\div3}} =\ \textgreater \ \boxed{\boxed{2a^2}}\\\\\\ H) \sqrt{64x^2y^8} = \sqrt{8^2x^2y^8}= \sqrt[2\div2]{8^{2\div2}\cdot x^{2\div2}\cdot y{8^{8\div2}}} =\ \textgreater \ \boxed{\boxed{8xy^4}}\\\\\\ I) \sqrt[4]{16x^8} = \sqrt[4]{2^4\cdot x^8} = \sqrt[4\div4]{2^{4\div4}\cdot x^{8\div2}} =\ \textgreater \ \boxed{\boxed{2x^4}}\\\\\\ J) \sqrt{25\cdot a^2\cdot b^2}= \sqrt{5^2\cdot a^2\cdot b^2}=\ \textgreater \ \boxed{\boxed{5ab }}\\\\\\ L) \sqrt{75} = \sqrt{5^2\cdot3}=\ \textgreater \ \boxed{\boxed{5\sqrt{3}}}$

$M) \sqrt{18}= \sqrt{3^2\cdot2}=\ \textgreater \ \boxed{\boxed{3\sqrt{2}}}\\\\\\ N) \sqrt{50} = \sqrt{5^2\cdot2}=\ \textgreater \ \boxed{\boxed{5\sqrt{2} }}\\\\\\ O) \sqrt{48^a}=\ \textgreater \ \boxed{\boxed{48^{ \frac{a}{2} } }}\\\\\\ P) \sqrt{ \frac{9}{x^2} } = \sqrt{\frac{3^2}{x^2}}=\ \textgreater \ \boxed{\boxed{\frac{3}{x} }}\\\\\\ Q) \sqrt{\frac{25a}{x^{10}}}=\sqrt{\frac{5^2\cdot a}{x^{10}}}= \sqrt[2\div2]{\frac{5^{2\div2}\cdot a^{1\div2}}{x^{10\div2}}}=\ \textgreater \ \boxed{\boxed{\frac{5a^{\frac{1}{2}}}{x^5}}}$

$R) \sqrt{\frac{49a^2}{16}}= \sqrt{\frac{7^2\cdot a^2}{4^2}}=\ \textgreater \ \boxed{\boxed{\frac{7a}{4}}}$

Acho que é só isso. rsrs
Espero ter ajudado.
Bons estudos!

Answer 2

Oie, Td Bom?!

A)

$\sqrt[8]{3 {}^{2} } = \sqrt[4]{3} = 3 {}^{ \frac{1}{4} }$

B)

$\sqrt{ {a}^{10} } = a {}^{5}$

C)

$\sqrt[4]{b} = b {}^{ \frac{1}{4} }$

D)

$\sqrt[]{ {16}^{} } = \sqrt{4 {}^{2} } = 4$

E)

$\sqrt[3]{64} = \sqrt[3]{4 {}^{3} } = 4$

F)

$\sqrt[3]{27} = \sqrt[3]{3 {}^{3} } = 3$

G)

$\sqrt[3]{8 {a}^{6} } = \sqrt[3]{8} \sqrt[3]{a {}^{6} } = \sqrt[3]{2 {}^{3} } a {}^{2} = 2a {}^{2}$

H)

$\sqrt{64x {}^{2} } {y}^{8} = \sqrt{64} \sqrt{x {}^{2} } y {}^{8} = \sqrt{8 {}^{2} } xy {}^{8} = 8xy {}^{8}$

I)

$\sqrt[4]{16x {}^{8} } = \sqrt[4]{16} \sqrt[4]{x {}^{8} } = \sqrt[4]{2 {}^{4} } x {}^{2} = 2x {}^{2}$

J)

$\sqrt{25 {a}^{2} } b {}^{2} = \sqrt{25} \sqrt{a {}^{2} } b {}^{2} = \sqrt{5 {}^{2} } ab {}^{2} = 5ab {}^{2}$

L)

$\sqrt{75} = \sqrt{5 {}^{2} \times 3} = \sqrt{5 {}^{2} } \sqrt{3} = 5 \sqrt{3}$

M)

$\sqrt{18} = \sqrt{3 {}^{2} \times 2 } = \sqrt{3 {}^{2} } \sqrt{2} = 3 \sqrt{2}$

N)

$\sqrt{50} = \sqrt{5 {}^{2} \times 2 } = \sqrt{5 {}^{2} } \sqrt{2} = 5 \sqrt{2}$

O)

$\sqrt{48 {}^{a} } = 48 {}^{ \frac{a}{2} }$

P)

$\sqrt{ \frac{9}{x {}^{2} } } = \frac{ \sqrt{9} }{ \sqrt{x {}^{2} } } = \frac{ \sqrt{3 {}^{2} } }{x} = \frac{3}{x}$

Q)

$\sqrt{ \frac{25 a}{ {x}^{10} } } = \frac{ \sqrt{5 {}^{2} \times a } }{x {}^{10} } = \frac{5a {}^{ \frac{1}{2} } }{x {}^{5} }$

R)

$\sqrt{ \frac{49 a {}^{2} }{16} } = \frac{ \sqrt{49a {}^{2} } }{ \sqrt{16} } = \frac{ \sqrt{49} \sqrt{a {}^{2} } }{ \sqrt{4 {}^{2} } } = \frac{ \sqrt{7 {}^{2} } a}{4} = \frac{7a}{4}$

Att. Makaveli1996