K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

12 tháng 11 2015

\(=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{x+2}{x+1}}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{x+1}{x+2}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2x+3}{x+2}}}}=\frac{1}{1+\frac{1}{1+\frac{x+2}{2x+3}}}=\frac{1}{1+\frac{1}{\frac{3x+5}{2x+3}}}=\frac{1}{1+\frac{2x+3}{3x+5}}=\frac{1}{\frac{5x+8}{3x+5}}=\frac{3x+5}{5x+8}\)

14 tháng 12 2019

Ta có:

M = \(\frac{1}{1-x}\cdot\frac{1}{1+x}\cdot\frac{1}{1+x^2}\cdot\frac{1}{1+x^4}\cdot\frac{1}{1+x^8}\cdot\frac{1}{1+x^{16}}\)

M = \(\frac{1}{\left(1-x\right)\left(1+x\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

M = \(\frac{1}{\left(1-x^2\right)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

M = \(\frac{1}{\left(1-x^4\right)\left(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\right)}\)

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

M = \(\frac{1}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)

M = \(\frac{1}{1-x^{32}}\)

14 tháng 12 2017

 = 1+x+1--x/1-x^2 +2/1+x^2+....+16/1+x^26

 = 2/1-x^2+2/1+x^2+....+16/1+x^16

 = ........

 = 16/1-x^16 + 16/1+x^16

 = 16+16x^16+16-16x^16/1-x^32

 = 32/1-x^32

k mk nha

ĐKXĐ: \(x\ne\pm1\)

\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

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

\(=\frac{32}{1-x^{32}}\)