\(\[\frac{1}{\sqrt{2}} +\frac{3}{\sqrt{8}} + \frac{6}{\sqrt{32}}.\] \)
Write in the form ,\(a \sqrt b\) where a and b are positive integers:
1/sqrt(2) + 3/sqrt(8) + 6/sqrt(32)
= (1sqrt(2) + 3sqrt(2) + 6sqrt(2))/sqrt(2)
= 10sqrt(2)/sqrt(2)
= 10sqrt(2^1)
= 10sqrt(2)
\(\frac{1}{\sqrt2}+\frac{3}{\sqrt{8}}+\frac{6}{\sqrt{32}}\\\\ \frac{1}{\sqrt2}+\frac{3}{2\sqrt{2}}+\frac{6}{4\sqrt{2}}\\\\ \frac{1}{\sqrt2}(1+\frac{3}{2}+\frac{6}{4})\\\\ \frac{\sqrt2}{2}4\\\\ 2\sqrt2 \)
+25 
