Matemática, perguntado por zybef, 9 meses atrás

DOU 250 PONTOS PRA QUEM RESPONDER CERTO

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​​

Soluções para a tarefa

Respondido por Usuário anônimo
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}


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Usuário anônimo: blz
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Respondido por Makaveli1996
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

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