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 = 2016

Sao đề lạ dữ vậy bạn kiểm tra lại xem cái phần B ấy

Đúng rồi bạn ạ

\(A=\frac{1+2+3+...+2015}{2016}\)

\(A=\frac{\left(2015+1\right)\times2015:2}{2016}\)

\(A=\frac{\text{2031120}}{2016}\)

\(A=\text{1007,5}\)

#)Giải :

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)

\(=0\)

\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)

\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)

=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)

\(=0\)

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 ~