\(\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}\)

\(=1+\frac{1}{2013}+1+\frac{1}{2012}+1+\frac{1}{2011}+1-\frac{3}{2014}\)

\(=4+\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2014}-\frac{1}{2014}-\frac{1}{2014}\right)\)

Ta có:

 \(\frac{1}{2011}>\frac{1}{2014}\Rightarrow\frac{1}{2011}-\frac{1}{2014}>0\)

\(\frac{1}{2012}>\frac{1}{2014}\Rightarrow\frac{1}{2012}-\frac{1}{2014}>0\)

\(\frac{1}{2013}>\frac{1}{2014}\Rightarrow\frac{1}{2013}-\frac{1}{2014}>0\)

\(\Rightarrow\frac{1}{2011}-\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2014}>0\)

\(\Rightarrow4+\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2014}-\frac{1}{2014}-\frac{1}{2014}\right)>4\)( thêm 2 vế với 4 )

\(\Rightarrow\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}>4\)

Vậy \(\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}>4\) 

Tham khảo nhé~

Mỗi số hạng của tổng đều nhỏ hơn 1 => Tổng đó nhỏ hơn 4

Ta có:

\(\frac{2014}{2013}+\frac{2013}{2012}+\frac{2012}{2011}+\frac{2011}{2014}=4+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{3}{2014}\)

Vì\(\frac{1}{2013}>\frac{1}{2014},\frac{1}{2012}>\frac{1}{2014},\frac{1}{2011}>\frac{1}{2014}\)

=>\(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}>\frac{3}{2014}\)

=>\(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{3}{2014}>0\)

=>\(4+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{3}{2014}>4\)

$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$

dễ ợt nhưng éo biết làm thông cảm nha

ban Dang Ha Trang an noi gi ki vay 

A = 2011x2017 = 2011x( 2014 +3) =2011x2014 + 2011 x3

B= 2014x2014 = 2014 x(2011 + 3) = 2014 x 2011 + 2014 x3 = 2011 x 2014 + 2014 x3

Vì 2011x3 < 2014x3 

=> A < B

A = (n + 2015)(n + 2016) + n² + n

= (n + 2015)(n + 2015 + 1) + n(n + 1)

Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2

=> (n + 2015)(n + 2015 + 1) chia hết cho 2

      n(n + 1) chia hết cho 2

=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2

=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)

\(A=\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2017}+\dfrac{2017}{2014}\\ =1-\dfrac{1}{2015}+1-\dfrac{1}{2016}+1-\dfrac{1}{2017}+1+\dfrac{1}{2014}+\dfrac{1}{2014}+\dfrac{1}{2014}\\ =\left(1+1+1+1\right)+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\\ =4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]\)

Vì \(\dfrac{1}{2015}< \dfrac{1}{2014}\), \(\dfrac{1}{2016}< \dfrac{1}{2014}\), \(\dfrac{1}{2017}< \dfrac{1}{2014}\)

\(\Rightarrow\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)< 0\\ \Rightarrow-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\\>0\\ \Rightarrow4+\left[-\left(\dfrac{1}{2015}-\dfrac{1}{2014}+\dfrac{1}{2016}-\dfrac{1}{2014}+\dfrac{1}{2017}-\dfrac{1}{2014}\right)\right]>4\)

Ta có: \(B=\frac{2011}{2012+2013+2014}+\frac{2012}{2012+2013+2014}+\frac{2013}{2012+2013+2014}\)

A= \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}\)

Xét từng số hạng của A và B

\(\frac{2011}{2012}>\frac{2011}{2012+2013+2014}\)

\(\frac{2012}{2013}>\frac{2012}{2012+2013+2014}\)

\(\frac{2013}{2014}>\frac{2013}{2012+2013+2014}\)

\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}>\frac{2011+2012+2013}{2012+2013+2014}\)

\(\Rightarrow A>B\)

Đề bạn ghi có hơi sai chút nên tự tự sửa lại nha!

chờ chút nhé bạn!

khó quá

\(A=\frac{2014^{2015}+2}{2014^{2016}+9}\)

\(2014A=\frac{2014\left(2014^{2015}+2\right)}{2014^{2016}+9}=\frac{2014^{2016}+4028}{2014^{2016}+9}=\frac{\left(2014^{2016}+9\right)+4019}{2014^{2016}+9}=\frac{2014^{2016}+9}{2014^{2016}+9}+\frac{4019}{2014^{2016}+9}=1+\frac{4019}{2014^{2016}+9}\)

\(B=\frac{2014^{2016}+2}{2014^{2017}+9}\)

\(2014B=\frac{2014\left(2014^{2016}+2\right)}{2014^{2017}+9}=\frac{2014^{2017}+4028}{2014^{2017}+9}=\frac{2014^{2017}+9+4019}{2014^{2017}+9}=\frac{2014^{2017}+9}{2014^{2017}+9}+\frac{4019}{2014^{2017}+9}=1+\frac{4019}{2014^{2017}+9}\)

Ta thấy:

\(2014^{2016}+9< 2014^{2017}+9\)

\(\Rightarrow\frac{4019}{2014^{2016}+9}>\frac{4019}{2014^{2017}+9}\)

\(\Rightarrow1+\frac{4019}{2014^{2016}+9}>1+\frac{4019}{2014^{2017}+9}\)

\(\Rightarrow A>B\)

Vậy ....