Question

ПОМОГИТЕ РЕШИТЬ, 9 ЗАДАНИЕ НА ФОТО

Answer 1

Ответ:

$x=0,01\\\\\displaystyle \frac{3+\, \dfrac{2}{3}\, x}{x^2+3x}+\frac{1}{3x+9}=\frac{9+2x}{3x(x+3)}+\frac{1}{3(x+3)}=\frac{9+2x+x}{3x(x+3)}=\frac{9+3x}{3x(x+3)}=\\\\\\=\frac{3(3+x)}{3x(x+3)}=\frac{1}{x}=\frac{1}{0,01}=\frac{1}{\frac{1}{100}}=100$