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\(\frac{2015+2016}{2016+2017}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
\(\frac{2015}{2016}>\frac{2015}{2016+2017}\)
\(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(A>B;\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
Tạm thời chỉ nghĩ ra được cách này -_-
Ta có :
\(A=\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2014}\)
\(A=\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2014+2}{2014}\)
\(A=\frac{2015}{2015}-\frac{1}{2015}+\frac{2016}{2016}-\frac{1}{2016}+\frac{2014}{2014}+\frac{2}{2014}\)
\(A=1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{2}{2014}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2015}+\frac{1}{2016}-\frac{2}{2014}\right)\)
\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]\)
Lại có :
\(\frac{1}{2015}< \frac{1}{2014}\)
\(\frac{1}{2016}< \frac{1}{2014}\)
\(\Rightarrow\)\(\frac{1}{2015}+\frac{1}{2016}< \frac{1}{2014}+\frac{1}{2014}\)
\(\Rightarrow\)\(\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)< 0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(\frac{1}{2014}+\frac{1}{2014}\right)\right]>3\)
Vậy \(A>3\)
Chúc bạn học tốt ~
Ta có:
\(B=\frac{2015+2016}{2016+2017}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
vì: \(\frac{2015}{2016}>\frac{2015}{2016+2017}\)VÀ \(\frac{2016}{2017}>\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
\(\Rightarrow\)\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{2015+2016}{2016+2017}\)
\(\Rightarrow A>B\)
Vậy: \(A>B\)
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
\(A=\frac{1+2+3+...+2015}{2016}\)
\(A=\frac{\left(2015+1\right)\times2015:2}{2016}\)
\(A=\frac{\text{2031120}}{2016}\)
\(A=\text{1007,5}\)