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VT
0
22 tháng 7 2016
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= \(\frac{x^{96}-1}{x-1}:\frac{x^{32}-1}{x-1}=\frac{x^{96}-1}{x^{32}-1}\)
21 tháng 6 2017
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 => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
21 tháng 6 2017
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 => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)
DN
2
21 tháng 2 2017
\(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)
\(\Rightarrow\left(\frac{x+1}{96}+1\right)+\left(\frac{x+2}{95}+1\right)=\left(\frac{x+3}{94}+1\right)+\left(\frac{x+4}{93}+1\right)\)
\(\Rightarrow\frac{x+97}{96}+\frac{x+97}{95}=\frac{x+97}{94}+\frac{x+97}{93}\)
\(\Rightarrow\frac{x+97}{96}+\frac{x+97}{95}-\frac{x+97}{94}-\frac{x+97}{93}=0\)
\(\Rightarrow\left(x+97\right)\left(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\right)=0\)
Mà \(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\ne0\)
\(\Rightarrow x+97=0\)
\(\Rightarrow x=-97\)
Vậy x = -97
D
21 tháng 2 2017
Có : \(\frac{x+1}{96}+\frac{x+2}{95}=\frac{x+3}{94}+\frac{x+4}{93}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{96}+1\right)+\left(\frac{x+2}{95}+1\right)\)= \(\left(\frac{x+3}{94}+1\right)+\left(\frac{x+4}{93}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}=\frac{x+97}{94}+\frac{x+97}{93}\)
\(\Leftrightarrow\) \(\frac{x+97}{96}+\frac{x+97}{95}-\frac{x+97}{94}-\frac{x+97}{93}=0\)
\(\Leftrightarrow\) \(\left(x+97\right)\left(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\right)=0\)
\(\Leftrightarrow\) \(\left(x+97\right)=0\) ( \(\frac{1}{96}+\frac{1}{95}-\frac{1}{94}-\frac{1}{93}\)) \(\ne0\)
\(\Leftrightarrow\)\(x=-97\)
Vậy \(x=-97\)
\(\frac{\left(x^{95}+x^{94}\right)+.....+\left(x+1\right)}{\left(x^{31}+x^{30}\right)+.....+\left(x+1\right)}=\frac{x^{94}\left(x+1\right)+......+\left(x+1\right)}{x^{30}\left(x+1\right)+.....+\left(x+1\right)}=\frac{x^{94}+x^{92}+....+x^2+1}{x^{30}+x^{28}+....+x^2+1}=\frac{\left(x^2+1\right)x^{92}+x^{88}\left(x^2+1\right).....+\left(x^2+1\right)}{\left(x^2+1\right)x^{28}+\left(x^2+1\right)x^{24}+....+\left(x^2+1\right)}=\frac{x^{92}+x^{88}+......+x^4+1}{x^{28}+x^{24}+.....+x^4+1}=\frac{x^{88}\left(x^4+1\right)+x^{80}\left(x^4+1\right)+....+\left(x^4+1\right)}{x^{24}\left(x^4+1\right)+x^{16}\left(x^4+1\right)+.....+\left(x^4+1\right)}=\frac{x^{88}+x^{80}+....+1}{x^{24}+x^{16}+...+1}\)
\(=\frac{x^{80}\left(x^8+1\right)+x^{64}\left(x^8+1\right)+.....+\left(x^8+1\right)}{x^{16}\left(x^8+1\right)+\left(x^8+1\right)}=\frac{x^{80}+x^{64}+.....+1}{x^{16}+1}=\frac{x^{64}\left(x^{16}+1\right)+.....+x^{16}+1}{x^{16}+1}=x^{64}+x^{32}+1\)