[tex]\displaystyle 1)\frac{x^2-x-110}{22+9x-x^2}=\frac{x^2+10x-11x-110}{-x^2+9x+22}=\frac{x(x+10)-11(x+10)}{-x^2+11x-2x+22}=\frac{(x+10)(x-11)}{-x(x-11)-2(x-11)}=\\\frac{(x+10)(x-11)}{-(x-11)(x+2)}=\frac{(x+10)(-1)}{x+2}=-\frac{x+10}{x+2}\\ \\2)(x^2-x-4)^2-10(x^2-x-4)+16=0\\ t^2-10t+16=0\\ t=2,t=8\\ x^2-x-4=2;x^2-x-4=8\\ x_1=-3,x_2=-2,x_3=3,x_4=4[/tex]