$<=>$$(1-\frac{2x-1}{x+1})^2[($ $(1-\frac{2x-1}{x+1})+6=$$\frac{12(2x-1)}{x+1}-20$
$=($$\frac{-x+2}{x+1)})^2($ $\frac{-x+2}{x+1}+6)=$ $\frac{4x-32}{x+1}$ $=$$\frac{(-x+2)^2}{(x+1)^2}.$ $\frac{5x+9}{x+1}=$ $\frac{4x-32}{x+1}$ $=$ $\frac{(-x+2)^2.(5x+9)}{(x+1)^2(x+1)}-$ $\frac{4x-32)}{x+1}=0$ $<=>$ $\frac{(-x+2)^2.(5x+9)-(4x-32)(x+1)^2}{(x+1)^3}=0$ $<=>(-x+2)^2(5x+9)-(4x-42)(x+1)^2=0$