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\(A\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)
\(A\left(x^2+x+1\right)=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)
(Giải thích: \(\left(x^2+x+1\right)\left(x^2-x+1\right)=\left(x^2+1\right)^2-x^2=x^4+x^2+1\))
\(A\left(x^2+x+1\right)=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)...\left(x^{32}-x^{16}+1\right)\)
.....
\(A\left(x^2+x+1\right)=x^{64}-x^{32}+1\)
\(\Rightarrow A=\frac{x^{64}-x^{32}+1}{x^2+x+1}\)
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\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
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a, \(\frac{x^{32}+x^{16}+1}{x^{16}+x^8+1}\)
\(=\frac{x^8+x^4+1}{x^4+x^2+1}\) Vậy phân thức \(a=\frac{x^8+x^4+1}{x^4+x^2+1}\)
P/s; Căn thức a, là phân số tối giản
b, \(\frac{x^8+3x^4+4}{x^4+x^2+2}\)
\(=\frac{x^4+3x^2+2}{x^2+x^1+1}\) Vậy căn thức \(b=\frac{x^4+3x^2+2}{x^2+x^1+1}\)
P/s; Căn thức b, có thể rút gọn được cho 2 và 4
Em ko chắc đâu nhé *-*
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-x^61+5*x^60+x^59-5*x^58-x^55+5*x^54+x^53-5*x^52-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^31+5*x^30+x^27-5*x^26-x^25+5*x^24+x^21-5*x^20-x^19+5*x^18+x^15-5*x^14-x^13+5*x^12+x^9-5*x^8-x^7+5*x^6+x^3-5*x^2-x+5
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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
= \(\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{16+16x^{16}+16-16x^{16}}{1-x^{32}}=\frac{32}{1-x^{32}}\)