Question

$\frac{5-b}{3-a} -\frac{16-b^2}{a^2-6a+9} *\frac{3-a}{b-4}$ буду благодарен за помощь)

Answer 1

$\frac{5-b}{3-a} - \frac{(4-b)(4+b)}{(a-3)^{2} } * \frac{3-a}{b-4} = \\= \frac{5-b}{3-a} - \frac{4+b}{a-3} = \frac{5-b}{3-a} + \frac{4+b}{3-a} = \frac{5-b+4+b}{3-a} = \frac{9}{3-a}$