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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) ta có: \(1-\frac{2016}{2017}=\frac{1}{2017}\)
\(1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow\frac{1}{2017}>\frac{1}{2018}\Rightarrow1-\frac{2016}{2017}>1-\frac{2017}{2018}\Rightarrow\frac{2016}{2017}< \frac{2017}{2018}\)
b) ta có: \(\frac{2017}{2016}-1=\frac{1}{2016};\frac{2018}{2017}-1=\frac{1}{2017}\)
\(\Rightarrow\frac{1}{2016}>\frac{1}{2017}\Rightarrow\frac{2017}{2016}-1>\frac{2018}{2017}-1\Rightarrow\frac{2017}{2016}>\frac{2018}{2017}\)
a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)
\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)
b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)
c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)
mà \(2017.2018< 2018.2019\)
\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)
\(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)
mà \(2017.2018+1< 2018.2019+1\)
\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)
\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)
\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)
ai lam nhanh ma dung minh h cho
cac ban nho giai ho minh nhe
\(\frac{2018}{\sqrt{2017}}+\frac{2017}{\sqrt{2018}}=\frac{2017}{\sqrt{2017}}+\frac{2018}{\sqrt{2018}}+\frac{1}{\sqrt{2017}}-\frac{1}{\sqrt{2018}}=\sqrt{2017}+\sqrt{2018}\)
\(\frac{1}{\sqrt{2017}}>\frac{1}{2018}\Rightarrow VT>\sqrt{2017}+\sqrt{2018}\)