Question

Please help!

If we write $\sqrt{2} + \frac{1}{\sqrt{2}} + \sqrt{3} + \frac{1}{\sqrt{3}}$ in the form $\frac{a \sqrt{2} + b \sqrt{3}}{c}$, such that $a$, $b$, $c$ are positive integers and $c$ is as small as possible then what is $a+b+c$.​​

Answer 1

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$\sqrt{2} + \frac{1}{ \sqrt{2} } + \sqrt{3} + \frac{1}{ \sqrt{3} } \\ \\ \sqrt{2} + \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } + \sqrt{3} + \frac{1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \sqrt{2} + \frac{ \sqrt{2} }{2} + \sqrt{3} + \frac{ \sqrt{3} }{3} \\ \\ \frac{6 \sqrt{2} + 3 \sqrt{2} + 6 \sqrt{3} + 2 \sqrt{3} }{6} \\ \\ \frac{9 \sqrt{2} + 8 \sqrt{3} }{6}$

$\frac{a \sqrt{2} + b \sqrt{3} }{6} \\ \\ a = 9 \\ b = 8 \\ c = 6 \\ \\ a + b + c \\ 9 + 8 + 6 \\ 23$

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Answer 2

A᭄ɴsᴡᴇʀ࿐ :−

\begin{gathered} \sqrt{2} + \frac{1}{ \sqrt{2} } + \sqrt{3} + \frac{1}{ \sqrt{3} } \\ \\ \sqrt{2} + \frac{1}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } + \sqrt{3} + \frac{1}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \sqrt{2} + \frac{ \sqrt{2} }{2} + \sqrt{3} + \frac{ \sqrt{3} }{3} \\ \\ \frac{6 \sqrt{2} + 3 \sqrt{2} + 6 \sqrt{3} + 2 \sqrt{3} }{6} \\ \\ \frac{9 \sqrt{2} + 8 \sqrt{3} }{6} \end{gathered}

2

+

2

1

+

3

+

3

1

2

+

2

1

×

2

2

+

3

+

3

1

×

3

3

2

+

2

2

+

3

+

3

3

6

6

2

+3

2

+6

3

+2

3

6

9

2

+8

3

\begin{gathered} \frac{a \sqrt{2} + b \sqrt{3} }{6} \\ \\ a = 9 \\ b = 8 \\ c = 6 \\ \\ a + b + c \\ 9 + 8 + 6 \\ 23\end{gathered}

6

a

2

+b

3

a=9

b=8

c=6

a+b+c

9+8+6

23

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