Question

Помогите решить, очень надо (если можно фотку на листочке )

Answer 1

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