3√8-2√12+√20----------------------- =3√18-2√27+√45

$\frac{3 \sqrt{8} -2 \sqrt{12}+ \sqrt{20} }{ 3\sqrt{18}-2 \sqrt{27}+ \sqrt{45}} = \frac{3 \sqrt{4*2}-2 \sqrt{4*3}+ \sqrt{4*5}}{3 \sqrt{9*2}-2 \sqrt{9*3}+ \sqrt{9*5}} = \frac{6 \sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}} = \\ \\ = \frac{2(3 \sqrt{2}-2\sqrt{3}+\sqrt{5})}{3(3\sqrt{2}-2\sqrt{3}+\sqrt{5})} = \frac{2}{3}$

Ответ: $\frac{2}{3}$