Express \(\dfrac{6}{\sqrt{245}+3\sqrt{125}+4\sqrt{5}}\) in the form where A and C are relatively prime integers, C is positive, and B is not divisible by the square of any prime. Find A+B+C .
+129852 Note that
sqrt (245) = sqrt ( 49 * 5) = 7sqrt (5) and
3sqrt (125) = 3 sqrt ( 25 * 5 ) = 15 sqrt (5)
So....the denominator is just 7sqrt (5) + 15sqrt (5) + 4sqrt (5) = 26sqrt (5)
So we have
6 3 sqrt (5) 3sqrt (5)
_________ = _________ _______ = ____________
26sqrt (5) 13sqrt (5) sqrt (5) 65
+592 \(\sqrt{245} = 7\sqrt{5}\)
\(3\sqrt{125} = 15\sqrt{5}\)
This means the denominator can be simplified to:
\((7 + 15 + 4)\sqrt{5} = 26\sqrt{5}\)
Which means the entire fraction is:
\(\frac{6}{26\sqrt{5}} = \frac{3}{13\sqrt{5}} = \frac{3\sqrt{5}}{65}\)
