Question

Найдите значение выражения
   (x-9) : (x^2-18x+81)/(x+9)
при x = 45

Answer 1

$\displaystyle (x-9):\frac{x^{2}-18x+81}{x+9}=(x-9)*\frac{x+9}{(x-9)^{2}}=\frac{x+9}{x-9};$

при х = 45:   $\displaystyle \frac{x+9}{x-9}= \frac{45+9}{45-9}= \frac{54}{36}= \frac{3}{2}=1,5$ 

Answer 2

$(x-9) : \frac{x^{2}-18x+81}{x+9} = (x-9) : \frac{(x-9)^{2}}{x+9} = \frac{(x-9)(x+9)}{(x-9)^{2}} = \frac{x+9}{x-9} \\ \\ x=45\\ \\ \frac{45+9}{45-9} = \frac{54}{36} =\frac{3}{2}$