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Cô nàng Thiên Yết
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zZz Cool Kid_new zZz
26 tháng 3 2020 lúc 13:39

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

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

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

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

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

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

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

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

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

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

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

Khách vãng lai đã xóa
Wendy
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kudo shinichi
27 tháng 11 2018 lúc 18:50

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

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

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

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

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

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

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

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

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

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

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

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

Anh Aries
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Trần Đức Thắng
6 tháng 8 2015 lúc 13:24

A = \(\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+...+\frac{16}{1+x^{16}}=\frac{2}{1-x^2}+\frac{2}{1+x^2}+..+\frac{16}{1+x^{16}}\)

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

Tưng tự 

\(\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\frac{32}{1-x^{32}}\)

Nguyễn Thảo Ngân
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Trần Thị Loan
3 tháng 7 2015 lúc 13:09

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

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

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

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

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

\(P=\frac{16}{1-x^{16}}+\frac{16}{x^{16}+1}=\frac{32}{\left(1-x^{16}\right)\left(1+x^{16}\right)}=\frac{32}{1-x^{32}}\)

Nguyễn Cao Đang
7 tháng 12 2017 lúc 11:21

toán lớp mấy?

Nguyễn Hoàng Anh Thư
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bùi huyền trang
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Hoàng Ngọc Tuyết Nhung
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nguyen van huy
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Nguyễn Anh Quân
14 tháng 12 2017 lúc 20:41

 = 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: \(x\ne\pm1\)

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

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

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

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

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

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

Bùi Hà Phương
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