Question

Найдите значение дробей

Answer 1

1)

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

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

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