Question

reduza os radicais e simplifique quando possivel
a)
$\sqrt[4]{ \sqrt{6 + \sqrt[3]{8} } }$
b)
$\frac{16}{ \sqrt{48} }$
resolva
a)
$\sqrt[3]{255} + \sqrt[3]{64} - \sqrt[4]{256}$
b)
$\sqrt[4]{1256} + \sqrt[5]{16} \times \sqrt[5]{32}$
c)
$(x { + 6)}^{2} = {x}^{2} - 3x$


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Answer 1

a)

$\sqrt[4]{ \sqrt{6 + \sqrt[3]{8} } }$

$\sqrt[4]{ \sqrt{6 + 2} } \\ \sqrt[8]{ 6 + 2 } \\ \sqrt[8]{8}$

b)

$\frac{16}{ \sqrt{48} }$

$\frac{16}{4 \sqrt{3} }$

$\frac{4}{ \sqrt{3} }$

$\frac{4}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$\frac{4 \sqrt{3} }{ \sqrt{3} \sqrt{3} }$

$\frac{4 \sqrt{3} }{3}$

A)

$\sqrt[3]{255} + \sqrt[3]{64} - \sqrt[4]{256} \\ \sqrt[3]{255} + 4 - 4 \\ \sqrt[3]{255} ou 6.34133$

B)

$\sqrt[4]{1256} + \sqrt[5]{16} \sqrt[5]{32} \\ \sqrt[4]{1256} + \sqrt[5]{16} \times 2 \\ \sqrt[4]{1256} + 2 \sqrt[5]{16}$

c)

${(x + 6)}^{2} = {x}^{2} - 3x$

${x}^{2} + 12x + 36 = {x}^{2} - 3x \\ 12x + 36 = - 3x \\ 12x + 3x = - 36$

$15x = - 36 \\ x = - \frac{36}{15}ou - \frac{12}{5} ou - 2.4$