Докажите равенство ( 2V2+3)/(3-2V2)=17+12V2

Решите задачу:

$\frac{2\sqrt{2}+3}{3-2\sqrt{2}}=\frac{(3+2\sqrt{2})(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}=\frac{(3+2\sqrt{2})^2}{3^2-(2\sqrt{2})^2}=\frac{3^2+2\cdot3\cdot2\sqrt{2}+(2\sqrt{2})^2}{9-2^2\cdot(\sqrt{2})^2}= \\ =\frac{9+12\sqrt{2}+2^2(\sqrt{2})^2}{9-2^2\cdot(\sqrt{2})^2}=\frac{9+12\sqrt{2}+4\cdot2}{9-4\cdot2}=\frac{9+12\sqrt{2}+8}{9-8}=\frac{17+12\sqrt{2}}{1}=17+12\sqrt{2}$