Question

Упростите выражение:

$frac{2y+1}{y^2+3y}+frac{y+2}{3y-y^2}-frac{1}{y}$

Answer 1

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

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

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

=1/y((2y^2+y-6y-3)/(y^2-9)-(y^2+2y+3y+6)/(y^2-9)-

-(y^2-9)/(y^2-9))=

=1/y((2y^2+y-6y-3-y^2-2y-3y-6-y^2+9)/(y^2-9))=

=1/y((2y^2-y^2-y^2+y-6y-2y-3y-3-6+9)/(y^2-9))=

=(-10)/(y^2-9)

Answer 2

$frac{2y+1}{y^2+3y}+frac{y+2}{3y-y^2}-frac{1}{y}=frac{2y+1}{y(y+3)}+frac{y+2}{y(3-y)}-frac{1}{y}=$

$=frac{(2y+1)(3-y)+(y+2)(y+3)-(y+3)(3-y)}{y(3-y)(y+3)}=$ 

$frac{6y+3-2y^2-y+y^2+2y+3y+6-9+y^2}{y(9-y^2)}=frac{10y}{y(9-y^2)}=frac{10}{9-y^2}$