Question

$\frac{15b}{5 - b} + \frac{6b}{b {}^{2} -25 } \times \frac{7b + 35}{3}$
Упростить выражение и решить по действиям. Срочно!!!! Пожалуйста.....

Answer 1

$\frac{15b}{5-b}+ \frac{6b}{b^2-25}* \frac{7b+35}{3}= \frac{15b}{-(b-5)}+ \frac{2b}{(b-5)*(b+5)}*7(b+5)= -\frac{15b}{b-5}+ \frac{2b}{b-5}*7$
$- \frac{15b}{b-5}+ \frac{14b}{b-5}=- \frac{b}{b-5}$

Answer 2

$\frac{15b}{5-b} + \frac{6b}{b^2-25} *\frac{7b+35}{3} =\frac{b}{5-b} \\ \\ 1)\frac{6b}{b^2-5^2}* \frac{7*b+7*5}{3}= \frac{6b}{(b-5)(b+5)}* \frac{7(b+5)}{3} = \frac{6b * 7(b+5)}{3*(b-5)(b+5)} = \\ \\ = \frac{3(b+5) * 2b*7}{3(b+5) * (b-5)} = \frac{14b}{b-5} \\ \\ 2) \frac{15b}{5-b} + \frac{14b}{b-5}= \frac{15b}{5-b} + \frac{14b}{-(5-b)} = \frac{15b }{5-b} + ( - \frac{14b}{5-b} ) = \frac{15b - 14b}{5-b} = \frac{b}{5-b}$