$$\frac{7+2\sqrt{3}}{7-2\sqrt{3}}$$
$\frac{28\sqrt{3}+61}{37}\approx 2.9593898$
$$\frac{\left(7+2\sqrt{3}\right)\left(7+2\sqrt{3}\right)}{\left(7-2\sqrt{3}\right)\left(7+2\sqrt{3}\right)}$$
$$\frac{\left(7+2\sqrt{3}\right)\left(7+2\sqrt{3}\right)}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{\left(7+2\sqrt{3}\right)^{2}}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{49+28\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{49+28\sqrt{3}+4\times 3}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{49+28\sqrt{3}+12}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{61+28\sqrt{3}}{7^{2}-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{61+28\sqrt{3}}{49-\left(-2\sqrt{3}\right)^{2}}$$
$$\frac{61+28\sqrt{3}}{49-\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}}$$
$$\frac{61+28\sqrt{3}}{49-4\left(\sqrt{3}\right)^{2}}$$
$$\frac{61+28\sqrt{3}}{49-4\times 3}$$
$$\frac{61+28\sqrt{3}}{49-12}$$
$$\frac{61+28\sqrt{3}}{37}$$
Show Solution
Hide Solution