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ta xét \(\frac{2016}{2017}+\frac{2017}{2018}=\frac{2016.2018}{2017.2018}+\frac{2017.2017}{2017.2018}\)
\(=\frac{2016.2018+2017.2017}{2017.2018}\)
Ta thấy \(2016+2017< 2016.2018+2017.2017\)
và \(2017+2018< 2017.2018\)
\(\Rightarrow\frac{2016+2017}{2017+2017}< \frac{2016}{2017}+\frac{2017}{2018}\)
lấy 2016+2017/2017+2018-2016/2017+2017/2018=0.(9)==>2016+2017/2017+2018>2016/2017+2017/2018
A=\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+2}+..........+\frac{2018}{2017^2+2017}\)
>\(\frac{2018}{2017^2+2017}+\frac{2018}{2017^2+2017}+........+\frac{2018}{2017^2+2017}\)
\(=\frac{2018}{2017^2+2017}.2017=\frac{2018.2017}{2017\left(2017+1\right)}=1\) (1)
Lại có:A<\(\frac{2018}{2017^2+1}+\frac{2018}{2017^2+1}+.........+\frac{2018}{2017^2+1}\)
\(=\frac{2018}{2017^2+1}.2017=\frac{2018.2017}{2017^2+1}=\frac{2017.\left(2017+1\right)}{2017^2+1}\)
\(=\frac{2017^2+2017}{2017^2+1}=\frac{2017^2+1+2016}{2017^2+1}=1+\frac{2016}{2017^2+1}< 2\) (2)
Từ (1) và (2) suy ra:1 < A < 2
Vậy A không phải là số nguyên
\(A=\frac{2016}{2017}+\frac{2017}{2018}+\frac{2018}{2019}\)
\(\Rightarrow A=(1-\frac{1}{2017})+(1-\frac{1}{2018})+(1-\frac{1}{2019})\)
\(\Rightarrow A=3-\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
\(\left(\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)<\(\frac{3}{2017}\)<\(1\)
\(\Rightarrow A\)>\(3-1=2\)
\(B=\frac{2016+2017+2018}{2017+2018+2019}\)
\(\Rightarrow B=1-\frac{3}{6054}\)
\(\Rightarrow B=1-\frac{1}{2018}\)
\(B\)<\(1\);\(A\)>\(2\)
\(\Rightarrow A\)>\(B\)
ai lam nhanh ma dung minh h cho
cac ban nho giai ho minh nhe
\(\frac{2018}{\sqrt{2017}}+\frac{2017}{\sqrt{2018}}=\frac{2017}{\sqrt{2017}}+\frac{2018}{\sqrt{2018}}+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}=\sqrt{2017}+\sqrt{2018}\)
\(\frac{1}{\sqrt{2017}}>\frac{1}{2018}\Rightarrow VT>\sqrt{2017}+\sqrt{2018}\)