Question

выполните действие и найдите значение данного выражения при b = √17​

Answer 1

$\dfrac{1}{b^2 - 6b + 9} - \dfrac{2}{b^2 - 9} + \dfrac{1}{9 + 6b + b^2} = \dfrac{1}{\left(b - 3\right)^2} - \dfrac{2}{(b-3)(b+3)} + \dfrac{1}{\left(b+3\right)^2} =\\\\\\= \dfrac{\left(b + 3\right)^2}{\left(b-3\right)^2\left(b + 3\right)^2} - \dfrac{2(b-3)(b+3)}{\left(b-3\right)^2\left(b + 3\right)^2} + \dfrac{\left(b-3\right)^2}{\left(b-3\right)^2\left(b + 3\right)^2} =\\\\\\= \dfrac{\left(b+3\right)^2 - 2(b-3)(b+3) + \left(b-3\right)^2}{\left(b-3\right)^2\left(b + 3\right)^2} =$

$=\dfrac{b^2 + 6b + 9 - 2\left(b^2 - 9\right) +b^2 - 6b + 9}{\left(b-3\right)^2\left(b + 3\right)^2} = \dfrac{2b^2 + 18 - 2b^2 + 18}{\left(b-3\right)^2\left(b + 3\right)^2} =\\\\\\= \dfrac{36}{{\left(b-3\right)^2\left(b + 3\right)^2}} = \dfrac{36}{\left(\left(b-3\right)\left(b+3\right)\right)^2} = \dfrac{36}{\left(b^2 - 9\right)^2} = \boxed{\left(\dfrac{6}{b^2 - 9}\right)^2}$

$\boldsymbol{b = \sqrt{17}}\, :$

$\left(\dfrac{6}{\left(\sqrt{17}\right)^2 - 9}\right)^2 = \left(\dfrac{6}{17 - 9}\right)^2 = \left(\dfrac{6}{8}\right)^2 = \left(\dfrac{3}{4}\right)^2 = \dfrac{3^2}{4^2} = \boxed{\boldsymbol{\dfrac{9}{16}}}$

Ответ:   $\dfrac{9}{16}$