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12 tháng 5 2019

\(A=\frac{2018^{2019}+1}{2018^{2019}-2017}=\frac{2018^{2019}-2017+2018}{2018^{2019}-2017}=\frac{2018^{2019}-2017}{2018^{2019}-2017}+\frac{2018}{2018^{2019}-2017}=1+\frac{2018}{2018^{2019}-2017}\)\(B=\frac{2018^{2019}+2}{2018^{2019}-2016}=\frac{2018^{2019}-2016+2018}{2018^{2019}-2016}=\frac{2018^{2019}-2016}{2018^{2019}-2016}+\frac{2018}{2018^{2019}-2016}=1+\frac{2018}{2018^{2019}-2016}\)Ta có: \(2018^{2019}-2017< 2018^{2019}-2016\)

\(\Rightarrow\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)

\(\Rightarrow1+\frac{2018}{2018^{2019}-2017}>1+\frac{2018}{2018^{2019}-2016}\)

\(\Rightarrow A>B\)

Vậy...

12 tháng 5 2019

Ta có :

\(A=\frac{2018^{2019}+1}{2018^{2019}-2017}=\frac{2018^{2019}-2017+2018}{2018^{2019}-2017}=1+\frac{2018}{2018^{2019}-2017}\)

\(B=\frac{2018^{2019}+2}{2018^{2019}-2016}=\frac{2018^{2019}-2016+2018}{2018^{2019}-2016}=1+\frac{2018}{2018^{2019}-2016}\)

Vì \(2018^{2019}-2017< 2018^{2019}-2016\)nên \(\frac{2018}{2018^{2019}-2017}>\frac{2018}{2018^{2019}-2016}\)hay \(A>B\)

~ Hok tốt ~

NV
7 tháng 5 2019

\(M=\left(2018+2018^2\right)+\left(2018^3+2018^4\right)+...+\left(2018^{2017}+2018^{2018}\right)\)

\(=2018\left(1+2018\right)+2018^3\left(1+2018\right)+...+2018^{2017}\left(1+2018\right)\)

\(=2018.2019+2018^3.2019+...+2018^{2017}.2019\)

\(=2019\left(2018+2018^3+...+2018^{2017}\right)⋮2019\)

b/ \(M=2018+2018^2+...+2018^{2018}\)

\(2018M=2018^2+2018^3+...+2018^{2018}+2018^{2019}\)

Lấy dưới trừ trên:

\(2018M-M=-2018+2018^{2019}\)

\(\Rightarrow2017M=2018^{2019}-2018\)

\(\Rightarrow M=\frac{2018^{2019}-2018}{2017}=\frac{2018^{2019}}{2017}-\frac{2017+1}{2017}=\frac{2018^{2019}}{2017}-1-\frac{1}{2017}\)

\(\Rightarrow M=N-\frac{1}{2017}\Rightarrow M< N\)

7 tháng 5 2019

Cảm ơn bạn đã giúp mình

1 tháng 9 2019

\(2^{2018}+2^{2017}>2^{2019}\)

nên \(2^{2018}+2^{2017}+...+2^2+2^1>2^{2019}\)

nên \(2^{2019}-\left(2^{2018}+2^{2017}+...+2^2+2^1\right)=\) số âm

số âm < 1

\(2^{2019}-\left(2^{2018}+2^{2017}+...+2^2+2^1\right)\)< 1

11 tháng 1 2020

\(A=2^{2015}+2^{2016}+2^{2017}+2^{2018}+2^{2019}+2^{2020}.\)

\(=2^{2014}\left(2+2^2+2^3+2^4+2^5+2^6\right)\)

\(=126.2^{2014}\)

\(=42.3.2^{2014}⋮42\)