Question

Calcule alguem me ajude a resolver todas essas contas
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Answer 1

$2. \frac{ \sqrt[6]{27} }{ \sqrt[4]{9} } = 2. \frac{ \sqrt[6]{ {3}^{3} } }{ \sqrt[4]{ {3}^{2} } } \\ = 2. \frac{ \sqrt{3} }{ \sqrt{3} } = 2.1 = 2$

$\frac{ \sqrt[3]{2}. \sqrt[5]{4} }{ \sqrt[10]{16} } = \frac{ \sqrt[3]{2}. \sqrt[5]{ {2}^{2} } }{ \sqrt[10]{ {2}^{4} } } = \frac{ \sqrt[3]{2}. \sqrt[5]{2 ^{2} } }{ \sqrt[5]{ {2}^{2} } } = \sqrt[3]{2}$

Decompondo os números 363, 27,12 e 192 em fatores primos teremos que

27|3. 12|2. 192|2. 363|3

9|3 6|2. 96|2. 121|11

3|3. 3|3. 48|2. 11|11

1. 1. 24|2. 1

27=3³ 12=2². 3. 12|2. 363=3.11²

6|2

3|3

1

$192 = {2}^{6} .3$

$\frac{ \sqrt{363} + \sqrt{27} + \sqrt{12} }{ \sqrt{192} } = \frac{ \sqrt{3. {11}^{2} } + \sqrt{ {3}^{2} .3} + \sqrt{ {2}^{2}.3 } }{ \sqrt{ {2}^{6}.3 } }$

$\frac{11 \sqrt{3} + 3 \sqrt{3} + 2 \sqrt{ 3} }{ {2}^{3} \sqrt{3} } = \frac{16 \sqrt{3} }{8 \sqrt{3} } = \frac{16}{2} = 8$