A = \(\left[\left(x^{95}+x^{94}+....+x^{64}\right)+\left(x^{63}+x^{62}+....+x^{32}\right)+\left(x^{31}+x^{30}+....+1\right)\right]:\left(x^{31}+x^{30}+....+1\right)\) Đặt thừa số chung

=> A = \(x^{64}+x^{32}+1\)

Chúc bạn làm bài tốt

Bạn ơi đặt thừa số chung thế nào vậy mình thấy có đặt được đâu

Ta có: TS= \(x^{95}+x^{94}+...+x+1\)(1)

=> x\(\cdot TS=x^{96}+x^{95}+...+x^2+x\)(2)

Từ (1)(2)=> \(\left(x-1\right)TS=x^{96}-1\)

=> \(TS=\frac{x^{96}-1}{x-1}\)

Ta có: MS=\(x^{31}+x^{30}+x^{29}+...+x+1\)(3)

=> x\(\cdot MS=x^{32}+x^{31}+x^{30}+...+x^2+x\)(4)

Từ (4)(3)=> \(\left(x-1\right)\cdot MS=x^{32}-1\)

<=> \(MS=\frac{x^{32}-1}{x-1}\)

Vậy A= \(\frac{x^{96}-1}{x-1}:\frac{x^{32}-1}{x-1}=\frac{x^{96}-1}{x^{32}-1}\)

Rút gọn.

\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)

ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Ta có:

\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)

\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Bài này từ 2 năm trước rồi mà

công nhận

\(\frac{\left(x^{95}+x^{94}\right)+.....+\left(x+1\right)}{\left(x^{31}+x^{30}\right)+.....+\left(x+1\right)}=\frac{x^{94}\left(x+1\right)+......+\left(x+1\right)}{x^{30}\left(x+1\right)+.....+\left(x+1\right)}=\frac{x^{94}+x^{92}+....+x^2+1}{x^{30}+x^{28}+....+x^2+1}=\frac{\left(x^2+1\right)x^{92}+x^{88}\left(x^2+1\right).....+\left(x^2+1\right)}{\left(x^2+1\right)x^{28}+\left(x^2+1\right)x^{24}+....+\left(x^2+1\right)}=\frac{x^{92}+x^{88}+......+x^4+1}{x^{28}+x^{24}+.....+x^4+1}=\frac{x^{88}\left(x^4+1\right)+x^{80}\left(x^4+1\right)+....+\left(x^4+1\right)}{x^{24}\left(x^4+1\right)+x^{16}\left(x^4+1\right)+.....+\left(x^4+1\right)}=\frac{x^{88}+x^{80}+....+1}{x^{24}+x^{16}+...+1}\)

\(=\frac{x^{80}\left(x^8+1\right)+x^{64}\left(x^8+1\right)+.....+\left(x^8+1\right)}{x^{16}\left(x^8+1\right)+\left(x^8+1\right)}=\frac{x^{80}+x^{64}+.....+1}{x^{16}+1}=\frac{x^{64}\left(x^{16}+1\right)+.....+x^{16}+1}{x^{16}+1}=x^{64}+x^{32}+1\)