If we write $\sqrt{5}+\frac{1}{\sqrt{5}} + \sqrt{7} + \frac{1}{\sqrt{7}}$ in the form $\dfrac{a\sqrt{5} + b\sqrt{7}}{c}$ such that $a$, $b$, and $c$ are positive integers and $c$ is as small as possible, then what is $a+b+c$?

Guest Jul 26, 2019

#1**+1 **

Simplify the following:

sqrt(5) + 1/sqrt(5) + sqrt(7) + 1/sqrt(7)

Rationalize the denominator. 1/sqrt(5) = 1/sqrt(5)×(sqrt(5))/(sqrt(5)) = (sqrt(5))/5:

sqrt(5) + (sqrt(5))/5 + sqrt(7) + 1/sqrt(7)

Rationalize the denominator. 1/sqrt(7) = 1/sqrt(7)×(sqrt(7))/(sqrt(7)) = (sqrt(7))/7:

sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7

Put each term in sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7 over the common denominator 35: sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7 = (35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35:

(35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35

(35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35 = (35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7))/35:

(35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7))/35

Add like terms. 35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7) = 42 sqrt(5) + 40 sqrt(7):

**(42 sqrt(5) + 40 sqrt(7))/35 = a + b + c =42 + 40 + 35 = 117**

Guest Jul 27, 2019