Vì \(B=\frac{2014^{11}+2}{2014^{12}+2}<1\)

\(\Rightarrow B=\frac{2014^{11}+2}{2014^{12}+2}<\frac{2014^{11}+2+4026}{2014^{12}+2+4026}=\frac{2014^{11}+4028}{2014^{12}+4028}=\frac{2014.\left(2014^{10}+2\right)}{2014\left(2014^{11}+2\right)}=\frac{2014^{10}+2}{2014^{11}+2}=A\)

Vậy B<A hay A<B

ta chứng minh bài toán phụ:

nếu ta có b<d \(\frac{a}{b}\)>\(\frac{c}{d}\) thì ad>bc

dễ thây \(\frac{ad}{bd}>\frac{cb}{bd}\)

 => ad>bd

áp dụng:

dat 2014=a ta co

\(A=\frac{a^{10}+2}{a^{11+2}}\)

 \(B=\frac{a^{11}+2}{a^{12}+2}\)

 ta có 

\(A=\frac{a^{10}+2.a^{12}+2}{a^{11}+2.a^{12}+2}\)

 \(B=\frac{a^{11}+2.a^{11}+2}{a^{12}+2.a^{11}+2}\)=\(\frac{a^{10}+2a^{12}+2}{a^{12}+2a^{11}+2}\)

 => A=B

mk hok chắc đâu nha

kb đc 0

2 câu đầu tôi làm đc

cùng nhân tử với 2014>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

B>A

\(\frac{2014^{2013}+1}{2014^{2013}-13}\)lớn hơn 1 là \(\frac{14}{2014^{2013}-13}\)

\(\frac{2014^{2012}+8}{2014^{2012}-11}\)lớn hơn 1 là \(\frac{19}{2014^{2012}-11}\)

\(\frac{14}{2014^{2013}-13}\)\(< \)\(\frac{19}{2014^{2012}-11}\)

\(\Rightarrow A< B\)

Ta có : A = \(\frac{2^{2014}+1}{2^{2014}}=1+\frac{1}{2^{2014}}\) 

           B = \(\frac{2^{2014}+2}{2^{2014}+1}=1+\frac{1}{2^{2014}+1}\)

Vì : \(\frac{1}{2^{2014}}>\frac{1}{2^{2014}+1}\)

Nên A > B 

Viết hẳn từng bước đi bạn

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

ta thấy:

2^2014<2^2014+2

=>\(\frac{2^{2014}+1}{2^{2014}}>\frac{2^{2014}+1}{2^{2014}+2}\)

vậy......

Có : 2²⁰¹⁴ + 1 > 2²⁰¹⁴ nên \(\frac{2^{2014}+1}{2^{2014}}\)> 1 .

2²¹⁰⁴ + 1 < 2²⁰¹⁴ + 2 nên \(\frac{2^{2014}+1}{2^{2014}+2}\)< 1.

=> \(\frac{2^{2014}+1}{2^{2014}}\)>\(\frac{2^{2014}+1}{2^{2014}+2}\)

Đặt :

\(A=\frac{2^{2014}+1}{2^{2014}}=\frac{2^{2014}}{2^{2014}}+\frac{1}{2^{2014}}=1+\frac{1}{2^{2014}}\)

\(B=\frac{2^{2014}+2}{2^{2014}+1}=\frac{2^{2014}+1+1}{2^{2014}+1}=\frac{2^{2014}+1}{2^{2014}+1}\)\(=1+\frac{1}{2^{2014}+1}\)

\(1+\frac{1}{2^{2014}}>1+\frac{1}{2^{2014}+2}\Leftrightarrow A>B\)