Question

ALGUEM ME RESPONDEEE PFV ;-;

Transforme em um so radical
a) ∜(5∛2)
b) ∛(2 ).∜2
c) \sqrt{2} x \sqrt[3]{2} x \sqrt[4]{2}

Answer 1

Propriedades:

$\boxed{\boxed{\sqrt[n]{a^{m}}=a^{m/n}}}\\\\\boxed{\boxed{\sqrt[n]{a}\cdot\sqrt[n]{b}\cdot\sqrt[n]{c}\cdot...\sqrt[n]{z}=\sqrt[n]{a\cdot b\cdot c\cdot...\cdot z}}}\\\\\boxed{\boxed{a^{x}\cdot a^{y}=a^{x+y}}}\\\\\boxed{\boxed{(a^{m})^{n}=(a^{n})^{m}=a^{m\cdot n}}}$
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a)

$\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{\sqrt[3]{5^{3}}\cdot\sqrt[3]{2}}\\\\\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{\sqrt[3]{125\cdot\sqrt[3]{2}}}\\\\\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{\sqrt[3]{125\cdot2}}\\\\\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{\sqrt[3]{250}}\\\\\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{\sqrt[3]{250^{1}}}\\\\\sqrt[4]{5\sqrt[3]{2}}=\sqrt[4]{250^{1/3}}\\\\\sqrt[4]{5\sqrt[3]{2}}=(250^{1/3})^{1/4}\\\\\sqrt[4]{5\sqrt[3]{2}}=250^{(1/3)\cdot(1/4)}\\\\\sqrt[4]{5\sqrt[3]{2}}=250^{1/12}$

$\sqrt[4]{5\sqrt[3]{2}}=\sqrt[12]{250^{1}}\\\\\boxed{\boxed{\sqrt[4]{5\sqrt[3]{2}}=\sqrt[12]{250}}}$

b)

$\sqrt[3]{2}\cdot\sqrt[4]{2}=\sqrt[3]{2^{1}}\cdot\sqrt[4]{2^{1}}\\\\\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{1/3}\cdot2^{1/4}\\\\\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{(1/3)+(1/4)}\\\\\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{(3+4)/(3\cdot4)}\\\\\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{7/12}\\\\\sqrt[3]{2}\cdot\sqrt[4]{2}=\sqrt[12]{2^{7}}\\\\\boxed{\boxed{\sqrt[3]{2}\cdot\sqrt[4]{2}=\sqrt[12]{128}}}$

c)

$\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{1/2}\cdot2^{1/3}\cdot2^{1/4}\\\\\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{(1/2)+(1/3)+(1/4)}$

O m.m.c entre 2, 3 e 4 é 12. Somando as frações:

$\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{(6+4+3)/12}\\\\\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}=2^{13/12}\\\\\boxed{\boxed{\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}=\sqrt[12]{2^{13}}}}$