Упростите выражение: 1/(b-3)-6b/(b^2-9)*(1/(b-2)-2/(b^2-2b)) и найдите его значение при b=1/2.

\[\frac{1}{b - 3} - \frac{6b}{b^{2} - 9} \cdot \left( \frac{1}{b - 2} - \frac{2}{b^{2} - 2b} \right) =\]

\[= \frac{1}{b + 3}\]

\[1)\ \frac{1^{\backslash b}}{b - 2} - \frac{2}{b^{2} - 2b} = \frac{b - 2}{b(b - 2)} = \frac{1}{b};\]

\[2)\frac{6b}{b^{2} - 9} \cdot \frac{1}{b} = \frac{6}{b^{2} - 9};\]

\[3)\frac{1^{\backslash b + 3}}{b - 3} - \frac{6}{b^{2} - 9} =\]

\[= \frac{b + 3 - 6}{(b - 3)(b + 3)} =\]

\[= \frac{b - 3}{(b - 3)(b + 3)} = \frac{1}{b + 3};\]

\[b = \frac{1}{2} = 0,5:\]

\[\frac{1}{0,5 + 3} = \frac{1}{3,5} = \frac{10}{35} = \frac{2}{7}.\]

\[Ответ:\ \frac{2}{7}.\]