Question

Срочно помогите даю 100 балов​

Answer 1

$\displaystyle\bf\\\frac{(b-6)^{2} }{2} \Big(\frac{1}{b^{2} -12b+36} +\frac{3}{36-b^{2} } \Big)+\frac{3b}{b+6} =\\\\\\=\frac{(b-6)^{2} }{2} \Big(\frac{1}{(b-6)^{2} } -\frac{3}{(b-6)(b+6)} \Big)+\frac{3b}{b+6} =\\\\\\=\frac{(b-6)^{2} }{2} \cdot\frac{b+6-3\cdot(b-6)}{(b-6)^{2} \cdot(b+6)} +\frac{3b}{b+6} =\\\\\\=\frac{(b-6)^{2} }{2} \cdot\frac{b+6-3b+18}{(b -6)^{2} \cdot(b+6)} +\frac{3b}{b+6} =\\\\\\=\frac{24-2b}{2\cdot(b+6)} +\frac{3b}{b+6} =\frac{12-b}{b+6} +\frac{3b}{b+6} =$

$\displaystyle\bf\\\frac{12-b+3b}{b+6} =\frac{2b+12}{b+6} =\frac{2\cdot(b+6)}{b+6} =2\\\\\\2=2$

Тождество доказано