Question

Пожалуйстаааааа оч срочно

Answer 1

(y*(y-3)/(y-3)^2-3*(y+3)/(y-3)*(y+3))*(1-3/y)=

=> (y-3/y-3)*(y-3/y)=1*(y-3/y)=y-3/y , при y=-3,6

=> -3,6-3/-3,6=6,6/3,6

Answer 2

$(\frac{y^{2}-3y }{y^{2}-6y+9 } -\frac{3y+9}{y^{2}-9 }) * (1 - \frac{3}{y} ) =$

$= (\frac{y(y-3) }{(y-3)^{2} } -\frac{3(y+3)}{(y-3)(y+3) } )* ( \frac{y-3}{y} ) =$

$= (\frac{y }{y-3 } -\frac{3}{y-3} )* ( \frac{y-3}{y} ) =$

$= \frac{y-3 }{y-3 } * \frac{y-3}{y} = 1 * \frac{y-3}{y} = \frac{y-3}{y}$

При y = -3,6

$\frac{-3,6-3}{-3,6} = \frac{-6,6}{-3,6} = \frac{6,6}{3,6} = 6 \frac{6}{10} : 3 \frac{6}{10} =$

$= \frac{66}{10} * \frac{10}{36} = \frac{66}{36} = \frac{11}{6} = 1 \frac{5}{6}$