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# need help

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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?

Jul 23, 2021

#1
+121096
+1

sqrt (2)  +  sqrt (3)  +   1  / [ (2sqrt (2)  + 3sqrt (3) ]

[(sqrt (2) + sqrt (3) ) ( 2sqrt ( 2)  + 3 sqrt (3) )   + 1 ]  /  ( 2sqrt (2) + 3sqrt (3) )

[  4  + 5sqrt (6)  +  9   +  1  ]  / ( sqrt (8)  + sqrt (27))

[ 14  + 5sqrt (6)  ]  / ( sqrt 8  +  sqrt (27)     mult num/den  by  sqrt (8)  - sqrt (27)

[14 + 5sqrt (6) [ sqrt (8)   - sqrt (27)]  / ( 8 - 27)

[14sqrt (8)  + 5 sqrt (48)  - 14sqrt(27)  - 5sqrt (162) ]  / (-19)

[ 14 * 2sqrt (2)  + 5*4 sqrt (3)  - 14*3 sqrt (3)  - 5*9 sqrt ( 2)  ]   / (-19)

[  -17 sqrt 2    -  22 sqrt 3  ]   /  (-19)

[ 17 sqrt 2   +   22 sqrt (3)   ]   /  19

a = 17     b  = 22     c   =19

a + b + c  =   58

Jul 23, 2021
#2
+26228
+1

If we write $$\sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) }$$
in the form $$\dfrac{(a*\sqrt{2} + b*\sqrt{3})}{c}$$ such that a, b, and c

are positive integers and c is as small as possible,
then what is a + b + c?

$$\begin{array}{|rcll|} \hline && \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{1}{ (2*\sqrt{2} + 3*\sqrt{3}) } * \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (2*\sqrt{2} + 3*\sqrt{3})(2*\sqrt{2} - 3*\sqrt{3}) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ (4*2-9*3) } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{(2*\sqrt{2} - 3*\sqrt{3})}{ -19 } \\\\ &=& \sqrt{2} + \sqrt{3} + \dfrac{-(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*(\sqrt{2} + \sqrt{3}) -(2*\sqrt{2} - 3*\sqrt{3})}{ 19 } \\\\ &=& \dfrac{19*\sqrt{2} + 19*\sqrt{3} -2*\sqrt{2} + 3*\sqrt{3}}{ 19 } \\\\ &=& \dfrac{{\color{red}17}*\sqrt{2} + {\color{red}22}*\sqrt{3}}{ {\color{red}19} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a+b+c &=& 17+22+19 \\ \mathbf{a+b+c} &=& \mathbf{58} \\ \hline \end{array}$$

Jul 24, 2021
edited by heureka  Jul 24, 2021