\frac{\frac{1}{4}\left(9+\sqrt{3}\right)}{\frac{1}{2}\left(\sqrt{3}+1\right)\left(\sqrt{\frac{3}{2}}\right)}

asked by guest
on Oct 24, 2024 at 11:26 pm



You asked:

Evaluate the expression: \(\frac{\frac{1}{4} \cdot \left(9 + \sqrt{3}\right)}{\frac{1}{2} \cdot \left(\sqrt{3} + 1\right) \cdot \sqrt{\frac{3}{2}}}\)

MathBot Answer:

\[\frac{\frac{1}{4} \cdot \left(9 + \sqrt{3}\right)}{\frac{1}{2} \cdot \left(\sqrt{3} + 1\right) \cdot \sqrt{\frac{3}{2}}} = \frac{\sqrt{6} \left(\sqrt{3} + 9\right)}{6 \cdot \left(1 + \sqrt{3}\right)} \approx 1.60368225335460104850473541106645\]

\[\begin{aligned}\frac{\frac{1}{4} \cdot \left(9 + \sqrt{3}\right)}{\frac{1}{2} \cdot \left(\sqrt{3} + 1\right) \cdot \sqrt{\frac{3}{2}}}& = \frac{1}{4} \cdot \left(9 + \sqrt{3}\right) \cdot \frac{1}{\frac{1}{2} \cdot \left(\sqrt{3} + 1\right) \cdot \sqrt{\frac{3}{2}}}\\& = \frac{1}{4} \cdot \left(9 + \sqrt{3}\right) \cdot \frac{1}{\frac{1}{2} \cdot \left(1 + \sqrt{3}\right) \cdot \sqrt{\frac{3}{2}}}\\& = \frac{1}{4} \cdot \left(9 + \sqrt{3}\right) \cdot \frac{1}{\frac{1}{2} \cdot \left(1 + \sqrt{3}\right) \cdot \frac{1}{2} \cdot \sqrt{6}}\\& = \frac{1}{4} \cdot \left(9 + \sqrt{3}\right) \cdot \frac{1}{\left(\frac{1}{2} + \frac{1}{2} \cdot \sqrt{3}\right) \cdot \frac{1}{2} \cdot \sqrt{6}}\\& = \frac{1}{4} \cdot \left(9 + \sqrt{3}\right) \cdot \frac{1}{\frac{1}{2} \cdot \sqrt{6} \cdot \left(\frac{1}{2} + \frac{1}{2} \cdot \sqrt{3}\right)}\\& = \frac{1}{4} \cdot \frac{1}{3} \cdot \sqrt{6} \cdot \frac{1}{\frac{1}{2} + \frac{1}{2} \cdot \sqrt{3}} \cdot \left(9 + \sqrt{3}\right)\\& = \frac{1}{12} \cdot \sqrt{6} \cdot \frac{1}{\frac{1}{2} + \frac{1}{2} \cdot \sqrt{3}} \cdot \left(9 + \sqrt{3}\right)\end{aligned}\]

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