Question

VALENDO 76 PONTOS



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​

Answer 1

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

Answer 2

Propriedades do radical aritmético

$\boxed{P_{1}} \mathsf{\sqrt[n]{a^n}=a}$

$\boxed{P_{2}}~\mathsf{\sqrt[n]{a^m}=\sqrt[n:p]{a^{m:p}}}$

$\boxed{P_{3}}\mathsf{\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}}$

$\dotfill$

$\boxed{P_{4}}\mathsf{\sqrt[n]{a.b}=\sqrt[n]{a}.\sqrt[n]{b}}$

a)

$\mathsf{\sqrt[8\div2]{3^{6\div3}}=\sqrt[4]{3^3}}$

b)

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

c) imcompleta

d)

$\mathsf{\sqrt{16}=\sqrt{2^2.2^2}=2.2=4}$

e)

$\mathsf{\sqrt[3]{64}=\sqrt[3]{(2^2)^3}=2^2=4}$

f)

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

g)

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

h)

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

i)

$\mathsf{\sqrt{64x^4y^8}=\sqrt{(2^3)^2.(x^2)^2.(y^4)^2}=8x^2y^4}$

j)

$\mathsf{\sqrt[4]{16x^3}=\sqrt[4]{2^4}.\sqrt[4\div2]{x^{2\div2}}.\sqrt[4]{x}=2\sqrt{x}.\sqrt[4]{x}}$

k)

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

l)

$\mathsf{\sqrt{12}=\sqrt{2^2.3}=2\sqrt{3}}$

m)

$\mathsf{\sqrt{75}=\sqrt{3.5^2}=5\sqrt{3}}$

n)

$\mathsf{\sqrt{18}=\sqrt{2.3^2}=3\sqrt{2}}$

o)

$\mathsf{\sqrt{50}=\sqrt{2.5^2}=5\sqrt{2}}$

p)

$\mathsf{\sqrt{12x}=\sqrt{2^2.3.x} =2\sqrt{3x}}$

q)

$\mathsf{\sqrt{48a}=\sqrt{(2^2)^2.3.a}=4\sqrt{3a}}$

r)

$\mathsf{\sqrt{\dfrac{9}{x^2}} =\dfrac{\sqrt{3^2}}{\sqrt{x^2}}=\dfrac{3}{x}}$

s)

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

t)

$\mathsf{\sqrt{\dfrac{49a^6}{16}}=\dfrac{\sqrt{7^2a^2}}{\sqrt{(2^2)^2}}=\dfrac{7a}{4}}$