\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2013}{2013}+\frac{1}{2013}+\frac{1}{2013}=\left(\frac{2013}{2014}+\frac{1}{2013}\right)+\left(\frac{2014}{2015}+\frac{1}{2013}\right)+1\)

Ta có: \(\frac{2013}{2014}+\frac{1}{2013}>\frac{2013}{2014}+\frac{1}{2014}=\frac{2014}{2014}=1\)

\(\frac{2014}{2015}+\frac{1}{2013}>\frac{2014}{2015}+\frac{1}{2015}=\frac{2015}{2015}=1\)

=> A > 1+ 1 + 1 = 3

\(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}=3,00000074\)

Vì 3,00000074 > 3 nên A > 3

a)\(\frac{2013}{2015}< \frac{2014}{2016}\)

b)\(\frac{2013+2014}{2014+2015}< \frac{2013}{2014}+\frac{2014}{2015}\)

ta có tính chất \(\frac{a}{b}\)>1 suy ra \(\frac{a.m}{b.m}\).........

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)

de ot la dau = nha

Ta có: \(\frac{2013}{2014}>\frac{2013}{2014+2015}\) (1)
\(\frac{2014}{2015}>\frac{2014}{2014+2015}\) (2)
ộng caác bất đẳng thứa (1) và (2) vào vế với vế:
\(\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\Rightarrow A>B\)

cảm ơn các bạn                         

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thì nhỏ hơn 3

tôi giảng cho bn nè nếu có 3 ps và đều tối giản nhưng chỉ có 1ps là ko lớn hơn 1 còn 2 ps kia thì lớn hơn 1 

=>3ps đó cộng vs nhau thì ko lớn hơn 3 vs dạng này

có \(\frac{2013}{2014}\)=1-\(\frac{1}{2014}\)

\(\frac{2014}{2015}\)=1-\(\frac{1}{2015}\)

\(\frac{2015}{2013}\)=1+\(\frac{2}{2013}\)

từ các ý trên suy ra A = 3 + \(\frac{2}{2013}\)-\(\frac{1}{2014}\)-\(\frac{1}{2015}\)=3+(\(\frac{1}{2013}\)-\(\frac{1}{2014}\))+(\(\frac{1}{2013}\)-\(\frac{1}{2015}\))

mặt khác \(\frac{1}{2013}\)>\(\frac{1}{2014}\);\(\frac{1}{2013}\)>\(\frac{1}{2015}\)suy ra  A>3

mỗi  số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15

ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

hay tong tren be hon 15

D\(\frac{2013}{2014+2015}+\frac{2014}{2014+2015}\)

Vì \(\frac{2013}{2014}>\frac{2013}{204+2015}\)

và \(\frac{2014}{2015}>\frac{2014}{2014+2015}\)

nên C>D

Ủng hộ mk nha

\(\frac{2013}{2014}+\frac{2014}{2015}=1,999...\)

\(\frac{2013+2014}{2014+2015}=4029\)

nen D>C