Question

Use propriedades dos radicais para achar um valor aproximado de:

$\sqrt{12} + \sqrt{32} =$
$\sqrt{20} + \sqrt{27} =$
​

Answer 1

Resposta:

Considerando:

$\sqrt{2} \approx1,41\\\sqrt{3} \approx1,73\\\sqrt{5} \approx2,24$

1) $\sqrt{12} + \sqrt{32} =$

Fatore:

12 | 2

6 | 2

3 | 3

1  | 1  /

= 2².3 = 12

32 | 2

16 | 2

8 | 2

4 | 2

2 | 2

1 | 1  /=$2^{5}$

$\sqrt{12} + \sqrt{32} = \sqrt{2^{2}.3}+\sqrt{2^{5}}=2^{\frac{2}{2}}.3^{\frac{1}{2} }+2^{\frac{5}{2}}=2.3^{\frac{1}{2} }+2^{\frac{4}{2}}.2^{\frac{1}{2}}=2\sqrt{3}+2^{2}.\sqrt{2}=2\sqrt{3}+4\sqrt{2}\approx 2.(1,73)+4.(1,41)\approx3,46+5,64\approx 9,06$

2) $\sqrt{20} + \sqrt{27} =$

​Fatore:

20 | 2

10 | 2

5 | 5

1 | 1

/=2².5

27 | 3

9 | 3

3 | 3

1 | 1/=$3^{3}=3^{2}.3$

$\sqrt{20} + \sqrt{27} =\sqrt{2^{2}.5} +\sqrt{3^{2}.3^{1}} =2\sqrt{5} +3\sqrt{3} \Rightarrow\\2\sqrt{5} +3\sqrt{3}\approx 2.(2,24)+3.(1,73)\approx9,67$

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Propriedades:

$(a^{m})^{n}=a^{m.n}\\\\\sqrt[n]{x^m} =x^{\frac{m}{n} }\\\\a^{m}a^{n}=a^{m+n}\\\\\frac{a^{m}}{a^{n}}=a^{m-n} \\\\a^{0}=1\\\\a^{1}=a$