$\frac{(-x^{2}-3x-2)}{x^{2}+4x+4}-\frac{2x^{2}-x-1}{x-1}\\\\ 1. \ -x^{2}-3x-2=0\\ D = b^{2}-4ac=(-3)^{2}-4\cdot (-1)\cdot (-2)=9-8=1\\ x_{1,2}=\frac{-b+/-\sqrt{D}}{2a}\\ x_{1}=\frac{3+1}{2\cdot (-1)}=\frac{4}{-2}=-2\\ x_{2}=\frac{3-1}{2\cdot (-1)}=\frac{2}{-2}=-1\\ -x^{2}-3x-2=-(x+1)(x+2)\\\\ 2. \ x^{2}+4x+4= (x+2)^{2}\\\\ 3. \ 2x^{2}-x-1=0\\ a+b+c = 0 \ => \ x_{1}=1, \ x_{2}=\frac{c}{a}\\ 2+(-1)+(-1)=0\\ x_{1}=1, \ x_{2}=\frac{-1}{2}=-\frac{1}{2}\\ 2x^{2}-x-1=2(x-1)(x+\frac{1}{2})\\\\$

$\frac{-(x+1)(x+2)}{(x+2)^{2}}-\frac{2(x-1)(x+\frac{1}{2})}{x-1}=-\frac{(x+1)(x+2)}{(x+2)^{2}}-\frac{2(x-1)(x+\frac{1}{2})}{x-1}=\\ =-\frac{x+1}{x+2}-2(x+\frac{1}{2}}) \ | \cdot (x+2)=-(x+1)-2(x+\frac{1}{2})(x+2)=\\ =-x-1-2(x^{2}+2x+\frac{1}{2}x+1)=-x-1-2(x^{2}+\frac{5}{2}+1)=\\ =-x-1-2x^{2}-5x-2=-2x^{2}-6x-3\\\\ -2x^{2}-6x-3=0\\ D = (-6)^{2}-4\cdot (-2)\cdot (-3)=36-24=12\\ x_{1}=\frac{6+\sqrt{12}}{-4}=-\frac{6+2\sqrt{3}}{4}=-\frac{2(3+\sqrt{3})}{4}=-\frac{3+\sqrt{3}}{2}\\$

$x_{2}=\frac{6-\sqrt{12}}{-4}=-\frac{6-2\sqrt{3}}{4}=-\frac{2(3-\sqrt{3})}{4}=-\frac{3-\sqrt{3}}{2}\\\\ -2x^{2}-6x-3=-2(x+\frac{3+\sqrt{3}}{2})(x+\frac{3-\sqrt{3}}{2})$