Question

URGENTEEEE

simplifique os radicais:
a) ⁸√3⁶
b) ¹⁵√a¹⁰
c) ⁴√⁸
d) √16
e) ³√64
f) ³√27
g) √25x²
h) ³√8
i) √64x⁴y⁸
j) ⁴√16x³
k) √25a²b⁴
l) √12
m) √75
n) √18
o) √50
p) √12x
q) √48a
r) √9/x²
s) √25a⁶/x¹⁰
t) √49a²/16


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Answer 1

Explicação passo-a-passo:

a) $\sqrt[8]{3^6}=\sqrt[8\div2]{3^{6\div2}}=\sqrt[4]{3^3}$

b) $\sqrt[15]{a^{10}}=\sqrt[15\div5]{a^{10\div5}}=\sqrt[3]{a^2}$

c) $\sqrt[4]{a^8}=\sqrt[4]{(a^2)^4}=a^2$

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) $\sqrt{25x^2}=\sqrt{(5x)^2}=5x$

h) $\sqrt[3]{8}=\sqrt[3]{2^3}=2$

i) $\sqrt{64x^4y^8}=\sqrt{(8x^2y^4)^2}=8x^2y^4$

j) $\sqrt[4]{16x^3}=\sqrt[4]{2^4\cdot x^3}=2\sqrt[4]{x^3}$

k) $\sqrt{25a^2b^4}=\sqrt{(5ab^2)}=5ab^2$

l) $\sqrt{12}=\sqrt{2^2\cdot3}=2\sqrt{3}$

m) $\sqrt{75}=\sqrt{3\cdot5^2}=5\sqrt{3}$

n) $\sqrt{18}=\sqrt{2\cdot3^2}=3\sqrt{2}$

o) $\sqrt{50}=\sqrt{2\cdot5^2}=5\sqrt{2}$

p) $\sqrt{12x}=\sqrt{2^2\cdot3x}=2\sqrt{3x}$

q) $\sqrt{48a}=\sqrt{(2^2)^2\cdot3a}=2^2\sqrt{3a}=4\sqrt{3a}$

r) $\sqrt{\dfrac{9}{x^2}}=\sqrt{\left(\dfrac{3}{x}\right)^2}=\dfrac{3}{x}$

s) $\sqrt{\dfrac{25a^6}{x^{10}}}=\sqrt{\left(\dfrac{5a^3}{x^5}\right)^2}=\dfrac{5a^3}{x^5}$

t) $\sqrt{\dfrac{49a^2}{16}}=\sqrt{\left(\dfrac{7a}{4}\right)^2}=\dfrac{7a}{4}$

Answer 2

Oie, Td Bom?!

a)

$\sqrt[8]{3 {}^{6} } = \sqrt[4]{3 {}^{3} } = \sqrt[4]{27}$

b)

$\sqrt[15]{a {}^{10} } = \sqrt[3]{a {}^{2} }$

c)

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

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)

$\sqrt{25x {}^{2} } = \sqrt{25} \sqrt{x {}^{2} } = \sqrt{5 {}^{2} } x = 5x$

h)

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

i)

$\sqrt{64x {}^{4} y {}^{8} } = \sqrt{64} \sqrt{x {}^{4} } \sqrt{y {}^{8} } = \sqrt{8 {}^{2} } x {}^{2} y {}^{4} = 8x {}^{2} y {}^{4}$

j)

$\sqrt[4]{16x {}^{3} } = \sqrt[4]{2 {}^{4} x {}^{3} } = \sqrt[4]{2 {}^{4} } \sqrt[4]{x {}^{3} } = 2 \sqrt[4]{x {}^{3} }$

k)

$\sqrt{25a {}^{2} b {}^{4} } = \sqrt{25} \sqrt{a {}^{2} } \sqrt{b {}^{4} } = \sqrt{5 {}^{2} } ab { }^{2} = 5ab {}^{2}$

l)

$\sqrt{12} = \sqrt{2 {}^{2} \times 3 } = \sqrt{2 {}^{2} } \sqrt{3} = 2 \sqrt{3}$

m)

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

n)

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

o)

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

p)

$\sqrt{12x} = \sqrt{2 {}^{2} \: . \: 3x } = \sqrt{2 {}^{2} } \sqrt{3x} = 2 \sqrt{3x}$

q)

$\sqrt{48a} = \sqrt{4 {}^{2} \: . \: 3a } = \sqrt{4 {}^{2} } \sqrt{3a} = 4 \sqrt{3a}$

r)

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

s)

$\sqrt{ \frac{25a {}^{6} }{x {}^{10} } } = \frac{ \sqrt{25a {}^{6} } }{ \sqrt{x {}^{10} } } = \frac{ \sqrt{25} \sqrt{a {}^{6} } }{x {}^{5} } = \frac{ \sqrt{5 {}^{2} }a {}^{3} }{x {}^{5} } = \frac{5a {}^{3} }{x {}^{5} }$

t)

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

Att. Makaveli1996