Question

Tính:

A= \(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

Answer 1

a) Ta có: \(A=\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}\)

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

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}\)

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

\(=\frac{16}{1-x^{16}}\)