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\(B=\left(2114-1\right)\left(2114+1\right)=2114^2-1< A=2114^2\)
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a) 11 1 4 − 2 5 7 + 5 1 4 = 11 1 4 − 2 5 7 − 5 1 4 = 11 1 4 − 5 1 4 − 2 5 7 = 6 − 2 5 7 = 5 7 7 − 2 5 7 = 3 2 7
b) 8 5 11 + 3 5 8 − 3 5 11 = 8 5 11 + 3 5 8 − 3 5 11 = 8 5 11 − 3 5 11 + 3 5 8 = 5 + 3 5 8 = 8 5 8
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a) 11 1 4 − 2 5 7 + 5 1 4 = 11 1 4 − 2 5 7 − 5 1 4 = 11 1 4 − 5 1 4 − 2 5 7 = 6 − 2 5 7 = 5 7 7 − 2 5 7 = 3 2 7
b) 8 5 11 + 3 5 8 − 3 5 11 = 8 5 11 + 3 5 8 − 3 5 11 = 8 5 11 − 3 5 11 + 3 5 8 = 5 + 3 5 8 = 8 5 8
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Ta có :
S = \(\dfrac{1-1}{1+2}.\dfrac{1-1}{1+2+3}.....\dfrac{1-1}{1+2+3+...+2114}\)
S = \(\dfrac{0}{1+2}.\dfrac{0}{1+2+3}....\dfrac{0}{1+2+3+...+2114}\)
S = 0.0....0
S = 0
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a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
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\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)
\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\)
\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)
Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\)
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a) ta có: \(1-\frac{2012}{2013}=\frac{1}{2013}\)
\(1-\frac{2013}{2014}=\frac{1}{2014}\)
mà \(\frac{1}{2013}>\frac{1}{2014}\) nên \(\frac{2013}{2014}>\frac{2012}{2013}\)
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2225 = (23)75 = 875
3151 > 3150 = (32)75 = 975
=> 3151 > 975 > 875
=> 3151 > 2225
4n - 5 chia hết cho 2n - 1
=> 4n - 2 - 3 chia hết cho 2n - 1
=> 2.(2n - 1) - 3 chia hết cho 2n - 1
Do 2.(2n - 1) chia hết cho 2n - 1 => 3 chia hết cho 2n - 1
Mà n thuộc N => 2n - 1 > hoặc = -1
=> 2n - 1 thuộc {-1 ; 1 ; 3}
=> 2n thuộc {0 ; 2 ; 4}
=> n thuộc {0 ; 1 ; 2}
3511>2114
3511>2111