Cho 1/(x-1)+1/(1+x)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)=

* Hướng dẫn giải

Ta có

$\begin{array}{l}\frac{1}{1-\mathrm{x}}+\frac{1}{1+\mathrm{x}}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\\ =\frac{1+\mathrm{x}+1-\mathrm{x}}{(1-\mathrm{x}\left)\right(1+\mathrm{x})}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\end{array}$

$\begin{array}{l}=\frac{2}{1-{\mathrm{x}}^2}+\frac{2}{1+{\mathrm{x}}^2}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\\ =\frac{2(1+{\mathrm{x}}^2)+2(1-{\mathrm{x}}^2)}{(1-{\mathrm{x}}^2\left)\right(1+{\mathrm{x}}^2)}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}\end{array}$

$\frac{4}{1-{\mathrm{x}}^4}+\frac{4}{1+{\mathrm{x}}^4}+\frac{8}{1+{\mathrm{x}}^8}=\frac{4(1+{\mathrm{x}}^4)+4(1-{\mathrm{x}}^4)}{(1-{\mathrm{x}}^4\left)\right(1+{\mathrm{x}}^4)}+\frac{8}{1+{\mathrm{x}}^8}$

$\frac{8}{1-{\mathrm{x}}^8}+\frac{8}{1+{\mathrm{x}}^8}=\frac{8(1+{\mathrm{x}}^8)+8(1-{\mathrm{x}}^8)}{(1-{\mathrm{x}}^8\left)\right(1+{\mathrm{x}}^8)}=\frac{16}{1-{\mathrm{x}}^{16}}$

Vậy số cần điền là 16.

Đáp án cần chọn là: A