+1224 \(\frac{1}{\sqrt{2}} + \frac{3}{\sqrt{8}}+\frac{5}{\sqrt{32}}\)
\(= \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} + \frac{3}{2\sqrt{2}} + \frac{5}{4\sqrt{2}}\)
\(= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{3}{2 \sqrt{2}} + \frac{\sqrt{2}}{\sqrt{2}} \cdot \frac{5}{4\sqrt{2}}\)
\(= \frac{\sqrt{2}}{2} + \frac{3 \sqrt{2}}{4} + \frac{5 \sqrt{2}}{8}\)
\(= \frac{4 \sqrt{2} + 6\sqrt{2} + 5\sqrt{2}}{8}\)
\(=\boxed{\frac{15\sqrt{2}}{8}}\)
.
+363 \(\frac{1}{\sqrt{2}}+\frac{3}{\sqrt{8}}+\frac{5}{\sqrt{32}}\)
\(\frac{1}{\sqrt{2}}+\frac{3}{2\sqrt{2}}+\frac{5}{4\sqrt{2}} \)
\(\frac{4}{4\sqrt{2}}+\frac{6}{4\sqrt{2}}+\frac{5}{4\sqrt{2}}\)
\(\frac{15}{4\sqrt{2}}\)
hmm........ we got different answers
+1224 These answers are equivalent, as shown below:
\(\frac{15}{4 \sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \boxed{\frac{15\sqrt{2}}{8}}\)
This process is called rationalizing the denominator.
