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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 = \(x^{64}+x^{32}+1\)
M(x)=x^95+x^94+x^93+.....+x^2+x+1
=x^64(x^31+x^30+...+x+1)+x^32(x^31+x^3... x^31+x^30+x^29+...+x^2+x+1
=(x^64+x^32+1)(x^31+x^30+x^29+...+x^2+...
=>dpcm
P(x) = M(x) * (x-1) = (x^96+x^95+x^94+ ...+x^2+x) - (x^95+x^94+ ...+x+1) = x^96-1
Q(x) = N(x) * (x-1) = (x^32+x^31+x^30+ ...+x^2+x) - (x^31+x^30+ ...+x+1) = x^32-1
Vì P(x) = x^96 - 1 = (x^32)^3 - 1 chia hết cho Q(x) (áp dụng hằng đẳng thức)
---> M(x) chia hết cho N(x) (đpcm)
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 => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
\(\frac{\left(x^{95}+x^{94}\right)+.....+\left(x+1\right)}{\left(x^{31}+x^{30}\right)+.....+\left(x+1\right)}=\frac{x^{94}\left(x+1\right)+......+\left(x+1\right)}{x^{30}\left(x+1\right)+.....+\left(x+1\right)}=\frac{x^{94}+x^{92}+....+x^2+1}{x^{30}+x^{28}+....+x^2+1}=\frac{\left(x^2+1\right)x^{92}+x^{88}\left(x^2+1\right).....+\left(x^2+1\right)}{\left(x^2+1\right)x^{28}+\left(x^2+1\right)x^{24}+....+\left(x^2+1\right)}=\frac{x^{92}+x^{88}+......+x^4+1}{x^{28}+x^{24}+.....+x^4+1}=\frac{x^{88}\left(x^4+1\right)+x^{80}\left(x^4+1\right)+....+\left(x^4+1\right)}{x^{24}\left(x^4+1\right)+x^{16}\left(x^4+1\right)+.....+\left(x^4+1\right)}=\frac{x^{88}+x^{80}+....+1}{x^{24}+x^{16}+...+1}\)
\(=\frac{x^{80}\left(x^8+1\right)+x^{64}\left(x^8+1\right)+.....+\left(x^8+1\right)}{x^{16}\left(x^8+1\right)+\left(x^8+1\right)}=\frac{x^{80}+x^{64}+.....+1}{x^{16}+1}=\frac{x^{64}\left(x^{16}+1\right)+.....+x^{16}+1}{x^{16}+1}=x^{64}+x^{32}+1\)
Bài này cậu hỏi lâu rồi nên không biết cậu muốn biết lời giải bài đó nữa không vậy?
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= \(\frac{x^{96}-1}{x-1}:\frac{x^{32}-1}{x-1}=\frac{x^{96}-1}{x^{32}-1}\)