Question

calcule as potencias
1 . √50 + 2√98 - √128 =
2. √2 . ³√2 - 6^√2^5 =
3. ³√-125 =

Answer 1

                                                                  ________
1 . √50 + 2√98 - √128 = √2.5² + 2√2.7² - √2².2².2².2

 5√2 + 2.7√2 - 2.2.2√2 ==>  5√2 + 14√2 - 8√2  ==> 11√2
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2. √2 . ³√2 - 6^√2^5 = 

6____   6___          6___   6___
√2³.2² - √2^5   ==>  √2^5 - √2^5  ===> 0
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3. ³√-125 = ∛-5³ ==> - 5 

Answer 2

$1) \\ \sqrt{50} +2 \sqrt{98} - \sqrt{128} = \\ \\ \sqrt{2.5^2} +2. \sqrt{2.7^2} - \sqrt{2^6.2} = \\ \\ 5 \sqrt{2} +2.7 \sqrt{2} -2^3 \sqrt{2} = \\ \\ 5 \sqrt{2} +14 \sqrt{2} -8 \sqrt{2} = \\ \\ 19 \sqrt{2} -8 \sqrt{2} = \\ \\ \fbox { 9 \sqrt{2} }$

$2) \\ \sqrt{2} . \sqrt[3]{2} - \sqrt[6]{2^5} = \\ \\ \sqrt[6]{2^3.2^2} - \sqrt[6]{2^5} = \\ \\ \sqrt[6]{2^5} - \sqrt[6]{2^5} = \\ \\ \fbox{ 0 }$

ao multiplicar raíz com índice diferente reduz ao mesmo com mmc

$3) \\ \sqrt[3]{-125} = \sqrt[3]{-5^3} =\fbox{ -5 }$