Sosánh: M= 5^2018+1/5^2017+1 và N=5^2017+1/5^2016+1
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Những câu hỏi liên quan
N
2
LN
1
19 tháng 7 2018
\(A=\frac{5^{2016}+1}{5^{2017}+1}\)
\(\Rightarrow5A=\frac{5^{2017}+5}{5^{2017}+1}=1+\frac{4}{5^{2017}+1}\)
\(B=\frac{5^{2017}+1}{5^{2018}+1}\)
\(\Rightarrow5B=\frac{5^{2018}+5}{5^{2018}+1}=1+\frac{4}{5^{2018}+1}\)
Do \(\frac{4}{5^{2018}+1}< \frac{4}{5^{2017}+1}\)
\(\Rightarrow5A>5B\Leftrightarrow A>B\)
NQ
0
N
22 tháng 8 2018
1: so sánh 2016/2017+2017/2018
vì 2016/2017 > 1/2017 >1/2018 =
> 2016/2017+2017/2018 >1/2018+2017/2018=1
vậy .....
LN
0
TM
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 12 2020
\(S=\dfrac{1}{2018!\left(2019-2018\right)!}+\dfrac{1}{2016!\left(2019-2016\right)!}+...+\dfrac{1}{2!\left(2019-2\right)!}+\dfrac{1}{0!\left(2019-0!\right)}\)
\(\Rightarrow2019!.S=\dfrac{2019!}{2018!\left(2019-2018\right)!}+\dfrac{2019!}{2016!\left(2019-2016\right)!}+...+\dfrac{2019!}{2!\left(2019-2\right)!}+\dfrac{2019!}{0!\left(2019-0\right)!}\)
\(=C_{2019}^{2018}+C_{2019}^{2016}+...+C_{2019}^2+C_{2019}^0\)
\(=\dfrac{1}{2}\left(C_{2019}^0+C_{2019}^1+...+C_{2019}^{2018}+C_{2019}^{2019}\right)\)
\(=\dfrac{1}{2}.2^{2019}=2^{2018}\)
\(\Rightarrow S=\dfrac{2^{2018}}{2019!}\)
`M=(5^2018+1)/(5^2017+1)`
`1/5M=(5^2017+1/5)/(5^2017+1)`
`1/5M=1-(4/5)/(5^2017+1)`
Tương tự:
`1/5N=1-(4/5)/(5^2016+1)`
`5^2017+1>5^2016+1`
`=>(4/5)/(5^2017+1)<(4/5)/(5^2016+1)`
`=>1-(4/5)/(5^2017+1)>1-(4/5)/(5^2016+1)`
`=>1/5M>1/5N=>M>N`
\(M=\dfrac{5^{2018}+1}{5^{2017}+1}=5-\dfrac{4}{5^{2017}+1}\)
\(N=\dfrac{5^{2017}+1}{5^{2016}+1}=5-\dfrac{4}{5^{2016}+1}\)
mà \(-\dfrac{4}{5^{2017}+1}>-\dfrac{4}{5^{2016}+1}\)
nên M>N