Câu này  cô làm rồi em nhá, em xem phần câu hỏi của tôi ý

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 *-*

\(=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+...+\dfrac{1}{x+2020}-\dfrac{1}{x+2022}\)

\(=\dfrac{x+2022-x}{x\left(x+2022\right)}=\dfrac{2022}{x\left(x+2022\right)}\)

\(=\dfrac{\left(x^{10}-x\right)+\left(x^5-x^2\right)+\left(x^2+x+1\right)}{x^8+x^4+1}\)

\(=\dfrac{x\left(x^9-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)}{x^8+2x^4+1-x^4}\)

\(=\dfrac{x\left(x^3-1\right)\left(x^6+x^3+1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)}{\left(x^4+1\right)^2-x^4}\)

\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x^7+x^4+x+x^2\right)+\left(x^2+x+1\right)}{\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)}\)

\(=\dfrac{\left(x^2+x+1\right)\left[\left(x-1\right)\left(x^7+x^2+x^4+x\right)+1\right]}{\left(x^4+2x^2+1-x^2\right)\left(x^4-x^2+1\right)}\)

\(=\dfrac{\left(x-1\right)\left(x^7+x^4+x^2+x\right)+1}{\left(x^2+1-x\right)\left(x^4-x^2+1\right)}\)

-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

\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

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

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

Ddeeff sao rồi bạn ko rút gọn được

\(\frac{x^{10}-x^8-x^7+x^6+x^6+x^4-x^3-x^2+1}{x^{30}+x^{24}+x^{18}+x^{12}+x^6+1}=\frac{(x^{10}-x^8+x^6)-(x^7-x^5+x^3)+(x^4-x^2+1)}{ (x^{30}+x^{18}+x^{24})+(x^{12}+x^6+1)} \)

=\(\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+x^6+1)(x^{18}+1 )}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{(x^{12}+2x^6+1-x^6) (x^6+1)(x^{12}-x^6+1)}=\frac{(x^4-x^2+1)(x^6-x^3+1)}{ (x^6-x^3+1)(x^6+x^3+1)(x^2+1)(x^4-x^2+1)(x^12-x^6+1 )} \)

=\(\frac{1}{(x^6+x^2+1)(x^2+1)(x^{12}-x^6+1)}\)