Question

только решить их нужно по действиям​

Answer 1

$1)\; \; \Big(\dfrac{2m+1}{2m-1}-\dfrac{2m-1}{2m+1}\Big):\dfrac{4m}{10\, m-5}=\dfrac{10\, m}{2m+1}\; ;\\\\\\a)\; \; \dfrac{2m+1}{2m-1}-\dfrac{2m-1}{2m+1}=\dfrac{(2m+1)^2-(2m-1)^2}{(2m+1)(2m-1)}=\dfrac{4m+4m}{(2m+1)(2m-1)}=\\\\\\=\dfrac{8m}{(2m+1)(2m-1)}\; ;\\\\\\b)\; \; \dfrac{8m}{(2m+1)(2m-1)}:\dfrac{4m}{10m-5}=\dfrac{8m}{(2m+1)(2m-1)}\cdot \dfrac{5\, (2m-1)}{4m}=\dfrac{10m}{2m+1}\; .$

$2)\; \; \Big(\dfrac{y+3}{y^2+9}\Big)\cdot \Big(\dfrac{y+3}{y-3}+\dfrac{y-3}{y+3}\Big)=\dfrac{2}{y-3}\; ;\\\\\\a)\; \; \dfrac{y+3}{y-3}+\dfrac{y-3}{y+3}=\dfrac{(y+3)^2+(y-3)^2}{(y+3)(y-3)}=\dfrac{2(y^2+9)}{(y+3)(y-3)}\; ;\\\\\\b)\; \; \dfrac{y+3}{y^2+9} \cdot \dfrac{2(y^2+9)}{(y+3)(y-3)}=\dfrac{2}{y-3}\; .$