Question

If √2 = 1.414, √3 = 1.732, √5 = 2.236 and √6 = 2.449, find the value of (2+√3 / 2-√3) + (2-√3 / 2+√3). + (√3-1 / √3+1)

Answer 1

$$\begin{gathered} \left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)+\left(\frac{2-\sqrt{3}}{2+\sqrt{3}}\right)+\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right) \\ =\frac{\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)}\times \frac{\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)}\times \frac{\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)}+\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}\times \frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)} \\ =\frac{4+2\sqrt{3}+2\sqrt{3}+3}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{4-2\sqrt{3}-2\sqrt{3}+3}{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}+\frac{3-\sqrt{3}-\sqrt{3}+1}{{\left(2\right)}^2-{\left(1\right)}^2} \\ =\frac{7+4\sqrt{3}}{4-3}+\frac{7-4\sqrt{3}}{4-3}+\frac{4-2\sqrt{3}}{3-1} \\ =\frac{7+4\sqrt{3}}{1}+\frac{7-4\sqrt{3}}{1}+\frac{4-2\sqrt{3}}{2} \\ =7+4\sqrt{3}+7-4\sqrt{3}+2-\sqrt{3} \\ =16-\sqrt{3} \\ =16-1.732 \\ =14.268 \end{gathered}$$