Найдите значение дробей

$\frac{3x}{x^3-3x^2}=\frac{3}{x^2-3x}$

ОДЗ: x≠0 и x≠3

$\frac{3}{(0,5)^2-3*0,5}=-\frac{12}{5}=-2,4\\\\\frac{3}{2^2-3*2}=-\frac{3}{2}=-1,5\\\\\frac{3}{(4,6)^2-3*4,6}=\frac{75}{184}$

2) $\frac{2c-3}{2c^3-3c^2}=\frac{2c-3}{c^2(2c-3)}=\frac{1}{c^2}$

ОДЗ: с≠0

$\frac{1}{(-2)^2}=\frac{1}{4}=0,25\\\\\frac{1}{4,5^2}=\frac{4}{81} \\\\\frac{1}{6,25^2}=\frac{16}{625}=0,0256\\\\ \frac{1}{\frac{2}{3}^2}=\frac{9}{4}=2.25\\\\ \frac{1}{1.5^2}=\frac{4}{9}$