Question

ПОЖАЛУЙСТА СРОЧНО ПОМОГИТЕ ДО ЗАВТРА !!!!!!!!!!!!!!!!!!

Answer 1

$\displaystyle\bf\\1)\\\\\frac{8}{1+\sqrt{3} } =\frac{8\cdot(1-\sqrt{3})}{(1+\sqrt{3})\cdot(1-\sqrt{3}) } =\frac{8\cdot(1-\sqrt{3})}{1^{2} -(\sqrt{3})^{2} } =\\\\\\=\frac{8\cdot(1-\sqrt{3})}{1-3 } =\frac{8\cdot(1-\sqrt{3})}{-2 } =-4(1-\sqrt{3})=4(\sqrt{3}-1)$

$\displaystyle\bf\\2)\\\\\Big(\frac{6}{y^{2}-9 }+\frac{1}{3-y} \Big)\cdot\frac{y^{2} +6y+9}{5} =\\\\\\= \Big(\frac{6}{(y-3)\cdot(y+3) }-\frac{1}{y-3} \Big)\cdot\frac{(y+3)^{2} }{5} = \\\\\\=\frac{6-y-3}{(y-3)\cdot(y+3)} \cdot\frac{(y+3)^{2} }{5} =\frac{3-y}{y-3} \cdot\frac{y+3}{5}=-\frac{y+3}{5}$