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If we write sqrt(2) + sqrt(3) + 1/(2*sqrt(2) + 3*sqrt(3)) in the form (a*sqrt(2) + b*sqrt(3))/c such that a, b, and c are positive integers and c is as small as possible, then what is a + b + c?

 Jan 30, 2022
 #1
avatar+514 
+3

The original expression is:

\(\sqrt{2} + \sqrt{3} + {1\over2\sqrt{2} + 3\sqrt{3}}\)

 

Next we can simplify the fraction part of the expression by multiplying the denominator by the difference of squares;

The expression is now:

\(\sqrt{2}+\sqrt{3} + {3\sqrt{3}-2\sqrt{2}\over19}\)

 

Then we can multiply the first two terms of the expression by \(19\) to get:

\({19\sqrt{2}\over19} + {19\sqrt{3}\over19} + {3\sqrt{3}-2\sqrt{2}\over19}\)

 

Then adding up the numerators and simplifying we get:

\(17\sqrt{2} + 22\sqrt{3}\over19\)

 

Thus, \(a\)\(b\), and \(c\) are \(17\)\(22\), and \(19\) respectively. 

 

Since we are looking for the sum of our three "variables", we add \(17+22+19\) to get:

\(58\), which is the answer.

 

:D

 Jan 31, 2022
 #2
avatar+360 
+3

\(\sqrt{2}+\sqrt{3}+\left(\frac{1}{2\sqrt{2}+3\sqrt{3}}\right)\)

\(\sqrt{2}+\sqrt{3}+\frac{1}{2\sqrt{2}+3\sqrt{3}}\)

\(\frac{\sqrt{2}\left(2\sqrt{2}+3\sqrt{3}\right)}{2\sqrt{2}+3\sqrt{3}}+\sqrt{3}+\frac{1}{2\sqrt{2}+3\sqrt{3}}\)

\(\frac{\sqrt{2}\left(2\sqrt{2}+3\sqrt{3}\right)}{2\sqrt{2}+3\sqrt{3}}+\frac{\sqrt{3}\left(2\sqrt{2}+3\sqrt{3}\right)}{2\sqrt{2}+3\sqrt{3}}+\frac{1}{2\sqrt{2}+3\sqrt{3}}\)

\(\frac{\sqrt{2}\left(2\sqrt{2}+3\sqrt{3}\right)+\sqrt{3}\left(2\sqrt{2}+3\sqrt{3}\right)+1}{2\sqrt{2}+3\sqrt{3}}\)

\(\frac{14+5\sqrt{6}}{2\sqrt{2}+3\sqrt{3}}\)

\(-\frac{-17\sqrt{2}-22\sqrt{3}}{19}\)

\(\frac{17\sqrt{2}+22\sqrt{3}}{19}\)

17+19+22=58, just the same as proyaop's answer!

 Jan 31, 2022

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