Question

3-b/b + b/3+b Подскажите решение!!!1

Answer 1

$\frac{3-b}{b} + \frac{b}{3+b} = \frac{(3-b)(3+b)+b^2}{b(3+b)}= \frac{3^2-b^2+b^2}{b(3+b)} = \frac{9}{b(3+b)}$