K
Khách

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

NV
22 tháng 2 2020

\(A\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)

\(A\left(x^2+x+1\right)=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)...\left(x^{32}-x^{16}+1\right)\)

(Giải thích: \(\left(x^2+x+1\right)\left(x^2-x+1\right)=\left(x^2+1\right)^2-x^2=x^4+x^2+1\))

\(A\left(x^2+x+1\right)=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)...\left(x^{32}-x^{16}+1\right)\)

.....

\(A\left(x^2+x+1\right)=x^{64}-x^{32}+1\)

\(\Rightarrow A=\frac{x^{64}-x^{32}+1}{x^2+x+1}\)

23 tháng 2 2020

Cảm ơn ạ.

13 tháng 11 2017

https://www.youtube.com/watch?v=cFZDEMTQQCs

j thế bạn

13 tháng 11 2015

h) \(=\frac{x+1}{32}\)

31 tháng 1 2016

câu hỏi tương tự

25 tháng 7 2023

\(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4=\left(\dfrac{3}{2}\right)^7\)

\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8=\left(\dfrac{1}{7}\right)^7\times2^{16}\)

\(\left(-\dfrac{1}{7}\right)^4\times125\times5=\left(-\dfrac{1}{7}\right)^4\times5^3\times5=\left(-\dfrac{1}{7}\right)^4\times5^4\)

\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:2^3:2^{-4}=2^0\)

\(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^3\times7^3=1^3\)
 

6, \(\dfrac{3}{2}\times\dfrac{9}{4}\times\dfrac{81}{16}=\dfrac{3}{2}\times\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^4\)

7,\(\left(\dfrac{1}{2}\right)^7\times8\times32\times2^8=\left(\dfrac{1}{2}\right)^7\times2^3\times2^5\times2^8\)

8,\(\left(-\dfrac{1}{7}\right) ^4\times125\times5=\left(\dfrac{1}{7}\right)^4\times5^3\times5\)

9,\(4\times32:\left(2^3\times\dfrac{1}{16}\right)=2^2\times2^5:\left[2^3\times\left(\dfrac{1}{2}\right)^4\right]\)

10, \(\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times49=\left(\dfrac{1}{7}\right)^2\times\dfrac{1}{7}\times7^2\)

6:=(3/2)*(3/2)^2*(3/2)^4=(3/2)^7

7: =(1/2)^7*2^3*2^5*2^8=2^9

8: =(-1/7)^4*5^4=(-5/7)^4

9: =2^2*2^5:(2^3/2^4)

=2^7/2=2^6

10: =(1/7)^3*7^2=1/7

26 tháng 3 2020

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

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

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

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

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

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

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

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

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

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

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