Question

номер 586 помогите пожалуйста,срочно нужно!

Answer 1

y/(3-y)+[(y²+3y)/(2y+3)]·[(3+y)/(y²-3y)-y/(y²-9y)]= 

=y/(3-y)+[y(y+3)/(2y+3)]·{[(3+y)(3+y)-y²]/[y(y-3)(y+3)]}=

=y/(3-y)+[1/(2y+3)]·{[(9+6y+y²)-y²]/[(y-3)]}=

=y/(3-y)+[1/(2y+3)]·{[(3(3+2y)]/[(y-3)]}=y/(3-y)+[1]·{[(3)]/[(y-3)]}=

=(y-3)/(3-y) =-1  не зависит от y/