$$\frac{1+\sqrt{2}}{3-2\sqrt{2}}$$
$5\sqrt{2}+7\approx 14.071067812$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{3^{2}-\left(-2\sqrt{2}\right)^{2}}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{9-\left(-2\sqrt{2}\right)^{2}}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{9-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{9-4\left(\sqrt{2}\right)^{2}}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{9-4\times 2}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{9-8}$$
$$\frac{\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)}{1}$$
$$\left(1+\sqrt{2}\right)\left(3+2\sqrt{2}\right)$$
$$3+2\sqrt{2}+3\sqrt{2}+2\left(\sqrt{2}\right)^{2}$$
$$3+5\sqrt{2}+2\left(\sqrt{2}\right)^{2}$$
$$3+5\sqrt{2}+2\times 2$$
$$3+5\sqrt{2}+4$$
$$7+5\sqrt{2}$$
Show Solution
Hide Solution
