Ta có:

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

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

\(\Rightarrow\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013+2014}{2014+2015}\)

\(\Rightarrow M>N\)

Ta có: \(N=\frac{2013+2014}{2014+2015}<1\);

          \(M=\frac{2013}{2014}+\frac{2014}{2015}>\frac{2013}{2015}+\frac{2014}{2015}=\frac{4027}{2015}>1\)

\(\Rightarrow A>B\)

TA CÓ 

2013>\(\frac{2013}{2014+2015}\) 

2014>\(\frac{2014}{2014+2015}\) 

=>2013+2014/2014+ 2015>2013+2014/2014+2015

=>M>N

\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)

Ta thấy: \(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\)

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

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

\(\Rightarrow M=\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}>N=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)

Vậy M>N

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 \(\frac{a.m}{b.m}\).........

Ta co: \(M=\frac{2013}{123456789}+\frac{2014}{987654321}=\frac{2013}{123456789}+\frac{2013}{987654321}+\frac{1}{987654321}\)

\(N=\frac{2013}{123456789}+\frac{1}{123456789}+\frac{2013}{987654321}\)

ma \(\frac{1}{987654321}< \frac{1}{123456789}\) nen \(M< N\)

\(M=\frac{2013}{123456789}+\frac{2014}{987654321}\)

\(N=\frac{2014}{123456789}+\frac{2013}{987654321}\)

\(M=\frac{2014}{987654321}-\frac{1}{987654321}\)

\(N=\frac{2014}{123456789}-\frac{1}{123456789}\)

Ta thấy \(\frac{1}{123456789}>\frac{1}{987654321}\)

\(\Rightarrow M< N\)

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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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)

Xét N có:

\(N=\frac{2012+2013+2014}{2013+2014+2015}=\frac{2012}{2013+2014+2015}+\frac{2013}{2013+2014+2015}+\frac{2014}{2013+2014+2015}\)

Ta các số hạng của M và N có:

\(\frac{2012}{2013}>\frac{2012}{2013+2014+2015}\) (1)

\(\frac{2013}{2014}>\frac{2013}{2013+2014+2015}\) (2)

\(\frac{2014}{2015}>\frac{2014}{2013+2014+2015}\) (3)

Từ (1);(2);(3) => M > N 

de ot la dau = nha

Gợi ý nhé: bạn hãy so sánh 2014A và 2014B rồi suy ngược lại A và B

Ta có:

2014A=2014²⁰¹⁴+ 2014/2014²⁰¹⁴+1=1+2013/2014²⁰¹⁴+1

2014B=2014²⁰¹³+2014/2014²⁰¹³+1=1+2013/2014²⁰¹³+1

vì 1+2013/2014²⁰¹⁴+1<1+2013/2014²⁰¹³+1 nên 10A < 10B

suy ra A<B