2((√3+√2)/(6√2))−1+3((√2+√3)/(4√3))−1
You asked:
MathBot Answer:
\[2 \cdot \frac{\sqrt{3} + \sqrt{2}}{6 \cdot \sqrt{2}} - 1 + 3 \cdot \frac{\sqrt{2} + \sqrt{3}}{4 \cdot \sqrt{3}} - 1 = -2 + \frac{\sqrt{2} \left(\sqrt{2} + \sqrt{3}\right)}{6} + \frac{\sqrt{3} \left(\sqrt{2} + \sqrt{3}\right)}{4} \approx 0.10395405949299087424886836446079\]
\[\begin{aligned}2 \cdot \frac{\sqrt{3} + \sqrt{2}}{6 \cdot \sqrt{2}} - 1 + 3 \cdot \frac{\sqrt{2} + \sqrt{3}}{4 \cdot \sqrt{3}} - 1& = 2 \cdot \left(\sqrt{3} + \sqrt{2}\right) \cdot \frac{1}{6 \cdot \sqrt{2}} - 1 + 3 \cdot \left(\sqrt{2} + \sqrt{3}\right) \cdot \frac{1}{4 \cdot \sqrt{3}} - 1\\& = 2 \cdot \left(\sqrt{2} + \sqrt{3}\right) \cdot \frac{1}{6 \cdot \sqrt{2}} - 1 + 3 \cdot \left(\sqrt{2} + \sqrt{3}\right) \cdot \frac{1}{4 \cdot \sqrt{3}} - 1\\& = 2 \cdot \frac{1}{12} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1 + 3 \cdot \left(\sqrt{2} + \sqrt{3}\right) \cdot \frac{1}{4 \cdot \sqrt{3}} - 1\\& = \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1 + 3 \cdot \left(\sqrt{2} + \sqrt{3}\right) \cdot \frac{1}{4 \cdot \sqrt{3}} - 1\\& = \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1 + 3 \cdot \frac{1}{12} \cdot \sqrt{3} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1\\& = \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1 + \frac{1}{4} \cdot \sqrt{3} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1\\& = \left(-1 + \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right)\right) + \frac{1}{4} \cdot \sqrt{3} \cdot \left(\sqrt{2} + \sqrt{3}\right) - 1\\& = \left(-1 + \frac{1}{4} \cdot \sqrt{3} \cdot \left(\sqrt{2} + \sqrt{3}\right) + \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right)\right) - 1\\& = -2 + \frac{1}{4} \cdot \sqrt{3} \cdot \left(\sqrt{2} + \sqrt{3}\right) + \frac{1}{6} \cdot \sqrt{2} \cdot \left(\sqrt{2} + \sqrt{3}\right)\end{aligned}\]