$\text{ If }\left|x\right|<1\text{ and }\frac{1-2x}{1-x+x^2}+\frac{2x-4x^3}{1-x^2+x^4}+\frac{4x^3-8x^7}{1-x^4+x^8}+---\mathrm{\infty }=\frac{1+ax}{1+bx+c{x}^2}\text{ then }a+b-c\text{ is \_\_\_\_\_}$

$\text{ Consider }\left(1-x+x^2\right)\left(1-x^2+x^4\right)\dots \dots \mathrm{\infty }$

$=\frac{\left(1+x+x^2\right)\left(1-x+x^2\right)\left(1-x^2+x^4\right)\dots \mathrm{\infty }}{\left(1+x+x^2\right)}$

$=\frac{1-x^{2^n}+x^{2^{n+1}}}{1+x+x^2}=\frac{1}{1+x+x^2}\left(\right|x|<1,n\to \mathrm{\infty })$

now take log on both sides and differentiate

$\log \left(1+x+x^2\right)+\log \left(1-x+x^2\right)+\log \left(1-x^2+x^4\right)+\log \left(1-x^4+x^8\right)+\cdots =0$

$\text{ Differentiating both sides w.r.t. }x\text{, we get }$

$\frac{1+2x}{1+x+x^2}+\frac{-1+2x}{1-x+x^2}+\frac{-2x+4x^3}{1-x^2+x^4}+\frac{-4x^3+8x^7}{1-x^4+x^8}+\cdots =0$

$\text{ or }\frac{1-2x}{1-x+x^2}+\frac{2x-4x^3}{1-x^2+x^4}+\frac{4x^3-8x^7}{1-x^4+x^8}+\cdots =\frac{1+2x}{1+x+x^2}$

$a=2,b=1,c=1$