+24 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 $a+b$?
+130081 \( $\frac{3}{\sqrt[5]{16}}+\frac{1}{\sqrt{3}}$ \)
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
