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1. Simplify $\frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}}$ and rationalize the denominator. The result can be expressed in the form $\frac{a^2\sqrt[5]{b}+b\sqrt{a}}{ab}$, where $a$ and $b$ are integers. What is the value of the sum $a+b$?

 Dec 27, 2019
 #1
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\( $\frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}}$ \)

 

Express in form

 

\($\frac{a^2\sqrt[5]{b}+b\sqrt{a}}{ab}$\)

 

 

Note that  (16)^(1/5)   =  (2^4)^(1/5)  = 2^(4/5)

So...multiply the numerator and denominator of the first fraction by 2^(1/5)

Multiply  the  numerator /denominator of the second fraction by  3^(1/2)

 

So  we have

 

3 * 2^(1/5)           3^(1/2)

__________  +    _______          get  a common denominator

       2                     3

 

3 * 3 * 2^(1/5)   + 2 * 3^(1/2)

_________________________       simplify

               2  *  3

 

3^2 *  5√ (2)      +   2 * √(3)

___________________________

                3 * 2

 

a =  3      b  = 2 

 

a +  b   =  5

 

 

cool cool cool

 Dec 27, 2019

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