Rationalize the denominator of: \(\frac{1}{\sqrt{2}+\sqrt{8}+\sqrt{32}}\). The answer can be written as \(\frac{\sqrt{A}}{B}\), where \(A\) and \(B\) are integers. Find the minimum possible value of \(A+B\).
The denominator of 1/(sqrt(2) + sqrt(8) + sqrt(32)) can be rationalized by multiplying both numerator and denominator by -sqrt(2) + sqrt(8) + sqrt(32). This gives us (-sqrt(2) + sqrt(8) + sqrt(32))/(-2 + 8 + 32) = sqrt(2)/19.
The final answer is 2 + 19 = 21.
+2666 \({1 \over \sqrt{2} + \sqrt{8} + \sqrt{32} }\)
\({1 \over \sqrt{2} + 2\sqrt{2} + 4\sqrt{2} }\)
\({1 \over 7\sqrt{2} }\)
\({1 \over 7\sqrt 2} \times { \sqrt{2} \over \sqrt {2}} \)
\({\sqrt 2 \over 14} \)
\(2 + 14 = \color{brown}\boxed{16}\)
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