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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 .

 Mar 1, 2021
 #1
avatar+118069 
+1

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

 

 

cool cool cool

 Mar 1, 2021
 #2
avatar+507 
+1

\(\sqrt{245} = 7\sqrt{5}\)

\(3\sqrt{125} = 15\sqrt{5}\)

 

This means the denominator can be simplified to:

\((7 + 15 + 4)\sqrt{5} = 26\sqrt{5}\)

 

Which means the entire fraction is:
\(\frac{6}{26\sqrt{5}} = \frac{3}{13\sqrt{5}} = \frac{3\sqrt{5}}{65}\)

 Mar 1, 2021

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