Question

помогите решить 5 номер слева

Answer 1

а)
$(\frac{b+1}{b-1}- \frac{b}{b+1} ): \frac{3b+1}{2b-2} \\ \\\frac{(b+1) ^{2}-b(b-1) }{(b-1)*(b+1)} * \frac{2b-2}{3b+1} \\ \\ \frac{(b+1)^{2}-b^{2}+b}{(b-1)*(b+1)} * \frac{2(b-1)}{3b+1} \\ \\ \frac{(b+1)^{2}-b^{2}+b}{(b-1)*(b+1)}* \frac{2(-1)}{3b+1}\\ \\ \frac{2((b+1)^{2}-b^{2}+b}{(b+1)*(3b+1)}\\ \\ \frac{2(b+1)^{2}-2b^{2}+2b}{(b+1)*(3b+1)} \\ \\ \frac{2((b+1)^{2}-b^{2}+b)}{(b+1)*(3b+1)}\\ \\ \frac{2(b^{2}+2b+1-b^{2}+b)}{(b+1)(3b+1)}\\ \\ \frac{2(3b+1)}{b+1)(3b+1)} \\ \\ \frac{2}{b+1} .$
  б)
$(\frac{c-2}{c+2}- \frac{c}{c-2})* \frac{c+2}{2-3c}\\ \\ \frac{(c-2)^{2}-c*(c+2)}{(c+2)*(c-2)} * \frac{c+2}{2-3c}\\ \\ \frac{(c-2)^{2}-c^{2}-2c}{c-2} * \frac{1}{2-3c} \\ \\ \frac{(c-2)^{2}-c^{2}-2c}{(c-2)*(2-3c)}\\ \\ \frac{c^{2}-4c+4-c^{2}-2c}{(c-2)*(2-3c)} \\ \\ \frac{-6c+4}{(c-2)*(2-3c)} \\ \\ \frac{2(-3c+2)}{(c-2)*(2-3c)} \\ \\ \frac{2}{c-2}$